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
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
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.
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.
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
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
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.
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.
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
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
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 lo-fi 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
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
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.
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.
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.
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.
We're only interested in the driver's pistonic (point source) BW, so can be considered a flat piston since the WLs are so large referenced to Sd.
GM
GM
Thanks guys (and gals).
I can arrive at this through calculations, a speaker cone, for lack of a better description, is a 'trapezoidal semi-cone'; meaning it is a cone with the pointy end chopped off.
So, if I determine the angle of the side of the cone relative to the base, I can then determine the entire cone area. Next, using the diameter of the dust cap, I can subtract the to 'pointy' area of the cone. Now, just add to that total, half the surface area of the surround.
But that seems like a lot of work, and it will certainly vary from speaker to speaker.
What I was hoping is that someone somewhere had already made these representative calculations, and had written it down. Again, I'm well aware that this will vary from speaker to speaker, but, as a rule of thumb, there must be some generalization of the relative cone area of the various size speakers.
Next, while I appreciate the effort, is the circular area of a speaker a fair approximation of the 'trapezoidal semi-cone' surface area? I really don't know, they may be proportional, and therefore a fair representative of the relative actual cone area.
My purpose in this, though I hate to admit it to this group, is in evaluating modern ready-make (oh the shame) speakers. In this day and age, the 1.4:1.0:0.7 standard box cabinets with large woofers seem to have gotten lost.
Now all we have are real, and apparent, exotic tower speakers with multi-woofers. Yet, I'm still an old fashioned guy who thinks in term of 10", 12", and 15" woofers. So, I'm looking for an admittedly superficial way of evaluating modern multi-woofer tower speakers in terms I can understand.
Are four 6.5" equivalent to two 8", are two 8" equivalent to one 12"? Again, this isn't necessarily about quality, one crappy 12" is certainly not better than two excellent 8". It's about perspective; it is about putting multi-woofer speakers into a perspective I can relate to.
I assumed, as I said, that someone somewhere had wondered the same thing, and had made a rough evaluation already.
Still, what you've given me is helpful.
Steve/Bluewizard
I can arrive at this through calculations, a speaker cone, for lack of a better description, is a 'trapezoidal semi-cone'; meaning it is a cone with the pointy end chopped off.
So, if I determine the angle of the side of the cone relative to the base, I can then determine the entire cone area. Next, using the diameter of the dust cap, I can subtract the to 'pointy' area of the cone. Now, just add to that total, half the surface area of the surround.
But that seems like a lot of work, and it will certainly vary from speaker to speaker.
What I was hoping is that someone somewhere had already made these representative calculations, and had written it down. Again, I'm well aware that this will vary from speaker to speaker, but, as a rule of thumb, there must be some generalization of the relative cone area of the various size speakers.
Next, while I appreciate the effort, is the circular area of a speaker a fair approximation of the 'trapezoidal semi-cone' surface area? I really don't know, they may be proportional, and therefore a fair representative of the relative actual cone area.
My purpose in this, though I hate to admit it to this group, is in evaluating modern ready-make (oh the shame) speakers. In this day and age, the 1.4:1.0:0.7 standard box cabinets with large woofers seem to have gotten lost.
Now all we have are real, and apparent, exotic tower speakers with multi-woofers. Yet, I'm still an old fashioned guy who thinks in term of 10", 12", and 15" woofers. So, I'm looking for an admittedly superficial way of evaluating modern multi-woofer tower speakers in terms I can understand.
Are four 6.5" equivalent to two 8", are two 8" equivalent to one 12"? Again, this isn't necessarily about quality, one crappy 12" is certainly not better than two excellent 8". It's about perspective; it is about putting multi-woofer speakers into a perspective I can relate to.
I assumed, as I said, that someone somewhere had wondered the same thing, and had made a rough evaluation already.
Still, what you've given me is helpful.
Steve/Bluewizard
Sorry Cal, perhaps I misunderstood. I really do appreciate the effort on your part, but I was under the impression that your calculations came for using circular area, hence my question about how proportional circular area and 'trapezoidal semi-cone' area were.
Maybe I confused your post with another. At any rate I do appreciate the effort and intend to use your calculations, they are very helpful.
Also, my last reply was to lend some sense of underlying purpose to my question. Again, I'm an old guy who thinks in old fashioned terms, and was looking for a way to convert my old mindset to modern terms.
So, using your calculations, if I divide the number derived for an 8" speaker into the number derived for a 12" speaker, I will have how many 8" it takes to make a 12"?
Ra = r12/r8 where 'R' or 'r' stand for Relative area.
Ra = 9.9/4.3 = 2.3
...meaning, generally and relatively speaking, it takes 2.3 8" speakers to equal one 12" speaker.
Continuing-
Ra = r10/r6.5 = 6.8/2.6 = 2.6
2.6 x 6.5" speakers is roughly equivalent to one 10" speaker.
Again, thanks, I do appreciate the effort.
Steve/bluewizard
Maybe I confused your post with another. At any rate I do appreciate the effort and intend to use your calculations, they are very helpful.
Also, my last reply was to lend some sense of underlying purpose to my question. Again, I'm an old guy who thinks in old fashioned terms, and was looking for a way to convert my old mindset to modern terms.
So, using your calculations, if I divide the number derived for an 8" speaker into the number derived for a 12" speaker, I will have how many 8" it takes to make a 12"?
Ra = r12/r8 where 'R' or 'r' stand for Relative area.
Ra = 9.9/4.3 = 2.3
...meaning, generally and relatively speaking, it takes 2.3 8" speakers to equal one 12" speaker.
Continuing-
Ra = r10/r6.5 = 6.8/2.6 = 2.6
2.6 x 6.5" speakers is roughly equivalent to one 10" speaker.
Again, thanks, I do appreciate the effort.
Steve/bluewizard
BlueWizard said:
Hmm, what part of this response didn't you understand?: http://www.diyaudio.com/forums/showthread.php?postid=1517220#post1517220
Calculating a diaphragm's surface area isn't required and neither is reducing it by the dustcap's area, so its effective diameter/area is all that's relevant, though you would for a fixed phase plug if you want better accuracy and of course you'd need to measure its air mass load to find its actual Sd for max accuracy.
Regardless, widest BW summation requires the drivers be butted up against each other, so as they are spread apart the point where they become discrete sources goes down in frequency requiring either one or more either be rolled off or all are XO limited.
GM
GM said:
Good explaination GM,
Weren't you part of the big debate on this over on the Bass List years ago? Surprisingly even some of the real experts had a problem with this concept. I may be wrong, but I "think" it was you and Dan Wiggins that were finally able to get this through to even some of the bone-heads like me😀
It's like an optical illustion that you can't see, until it's pointed out to you. Afterwards, you can't see anything else but the picture. Then you go crazy trying to figure out why you couldn't see it at first.
Best Regards,
TerryO
Greets!
Thanks!
Yeah, some things in audio are a bit counter-intuitive and this is one of them. Don't recall, but it's a subject I would have 'dug my heels in' on though. Hu Bath probably was one as he seemed to know a lot about everything (wonder what's happened to him) and Ron E. if it was later on.
Yeah, I hate it when I can't see the 'forest for the trees'. During my 'adventures' in horn design/building, I waltzed all around Unity concept/tapped bass horns with focussed MTMs and true corner loaded pipe horns, but never quite put two and two together the way TD did until he 'teased' me with a description of his prototype back in a late '90s PM. Talk about feeling like an idiot........ 🙁 Oh well, that's why he's a successful inventor and I'm not.
GM
Thanks!
Yeah, some things in audio are a bit counter-intuitive and this is one of them. Don't recall, but it's a subject I would have 'dug my heels in' on though. Hu Bath probably was one as he seemed to know a lot about everything (wonder what's happened to him) and Ron E. if it was later on.
Yeah, I hate it when I can't see the 'forest for the trees'. During my 'adventures' in horn design/building, I waltzed all around Unity concept/tapped bass horns with focussed MTMs and true corner loaded pipe horns, but never quite put two and two together the way TD did until he 'teased' me with a description of his prototype back in a late '90s PM. Talk about feeling like an idiot........ 🙁 Oh well, that's why he's a successful inventor and I'm not.
GM
We're only interested in the driver's pistonic (point source) BW, so can be considered a flat piston since the WLs are so large referenced to Sd.
Well, in all honesty, it would have made more sense if you had defined BW, WL, and Sd.
Still, I get your point. I did read it and, to an extent, understand it, but I wasn't sure what Cal based his figures on.
But this brings up another question, which in a sense, was the reason for the original question.
Why is the market dominated by multi-woofer tower designs?
You said yourself, if I interpret correctly, that to make this equivalence between several small and one big speaker, the small speakers need to be clustered very close together. But that is not how they are configured in common tower designs.
My design books, such as they are, tell me that that is the absolute worst possible configuration of speakers for musical reproduction. Yet, again, it is a design that dominates the market.
I would suspect you would be extremely hard pressed to find a consumer 1.4:1.0:0.7 classic single woofer design out there. The only ones I know of are a minor low-end line of Yamaha and Cerwin Vaga's home audio line, and I doubt that either of those speaker lines have made great headway into the consumer market.
Even here, I think you will find few people designing 'classic' box single big woofer speaker designs.
So, thank to all for the help. I did get my core question answered.
Steve/bluewizard
Well, in all honesty, it would have made more sense if you had defined BW, WL, and Sd.
Still, I get your point. I did read it and, to an extent, understand it, but I wasn't sure what Cal based his figures on.
But this brings up another question, which in a sense, was the reason for the original question.
Why is the market dominated by multi-woofer tower designs?
You said yourself, if I interpret correctly, that to make this equivalence between several small and one big speaker, the small speakers need to be clustered very close together. But that is not how they are configured in common tower designs.
My design books, such as they are, tell me that that is the absolute worst possible configuration of speakers for musical reproduction. Yet, again, it is a design that dominates the market.
I would suspect you would be extremely hard pressed to find a consumer 1.4:1.0:0.7 classic single woofer design out there. The only ones I know of are a minor low-end line of Yamaha and Cerwin Vaga's home audio line, and I doubt that either of those speaker lines have made great headway into the consumer market.
Even here, I think you will find few people designing 'classic' box single big woofer speaker designs.
So, thank to all for the help. I did get my core question answered.
Steve/bluewizard
BlueWizard said:
Hi Steve,
I simply measured the speaker cone area and reduced the 8.3 in^2 of the 4.5" driver to 1 and made them all relative. Here is a list of the actual measurements.
4.5" = 8.3 in^2
5.0 = 11.0
6.5 = 21.6
8.0 = 35.7
10 = 56.7
12 = 82.5
15 = 132.7
Again, keep in mind these areas are approximate and reflect only the drivers I measured. Different manufacturers have different ways of measuring the speaker size. For example the 5" MCM drivers are 5" outside to outside on the basket where as the 8" TB is the centre to centre of the bolt pattern. The frame is actually 8 5/16"
Anyway, the others have filled you in on why this isn't really that important in the scheme of things unless you plan to design around piston size.
Why is the market dominated by multiple woofer designs? Lots of reasons I suppose, including power handling but I think the number one reason is...
...they sell better and don't cost a whole lot more to produce..
For low frequencies:
What matters with respect to maximum output level is the volume displacement. The max volume can be calculated as Vmax=Sd*Xmax*n. Sd is the equivalient piston area, Xmax is the maximum displacement of the cone and n is the number of drivers.
Sd is NOT calculated as the area of the cone, but as the area of an imaginary circular disc with a diameter corresponding to the driver diameter (measured at the middle of the cone suspension).
Xmax is the maximum cone excursion without severe distortion, whatever that is.
The maximum sound pressure in free space can be calculated as
pmax=(w²*Vmax*rho0)/(4*pi*r)
where w=2*pi*frequency, rho0=1.2kg/m³, and r equals distance between source and listener.
and the maximum sound pressure level, in dB, can be calculated as
pmax=20*log10(pmax/pref)
where pref=20 µPa, and Xmax, Vmax and pmax are RMS values, ie peak value/sqrt(2)
For high frequencies it is more complicated, but then again multiple drivers are most commonly used for low frequencies 😀 .
What matters with respect to maximum output level is the volume displacement. The max volume can be calculated as Vmax=Sd*Xmax*n. Sd is the equivalient piston area, Xmax is the maximum displacement of the cone and n is the number of drivers.
Sd is NOT calculated as the area of the cone, but as the area of an imaginary circular disc with a diameter corresponding to the driver diameter (measured at the middle of the cone suspension).
Xmax is the maximum cone excursion without severe distortion, whatever that is.
The maximum sound pressure in free space can be calculated as
pmax=(w²*Vmax*rho0)/(4*pi*r)
where w=2*pi*frequency, rho0=1.2kg/m³, and r equals distance between source and listener.
and the maximum sound pressure level, in dB, can be calculated as
pmax=20*log10(pmax/pref)
where pref=20 µPa, and Xmax, Vmax and pmax are RMS values, ie peak value/sqrt(2)
For high frequencies it is more complicated, but then again multiple drivers are most commonly used for low frequencies 😀 .
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