Cone area

[basscort2009 5,000+ posts]

hes tryin to do less room though. so 1 ten would take very small room and be loud as hell with a big *** amp on it

 

[michaellane 5,000+ posts]

ya id like to try two beef 10's off like 1500 a peice or so...a guy on here has a accord and runs two hdc3 10's off a subdown 1500 and got a 145 i beleive..

 

[basscort2009 5,000+ posts]

not a bad score with some tens

 

[michaellane 5,000+ posts]

ya i hope mine is around that. im doing a new box sunday and hope to get metered sometime this week... //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[Bobby322 10+ year member]

thanks for the help guys. I probally will do 3 10''s if I can find some.

I want to keep the same amp I have now, does 2400w @ 1 ohm.

 

[Compjake 10+ year member]

4x 8" or 3x 10" will be more cone are than a single 15" sub.

Yes, I did see that. //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[snb778 5,000+ posts]

3x 10's to a 18"

 

[slam 5,000+ posts]

just get 1 l7 8 and do a 144 like me //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[audioholic 5,000+ posts]

this is wrong, that would be under the assumption that the subwoofer actually had a piston diameter that was equal to the outside of the surround, not the inside.
the number the OP is looking for when the term "cone-area" is used is effective piston area, basically its how much of the sub is actually moving air. However, 85% of the people on here go by the method previously mentioned which is WRONG

some average values would be

8" = 195cm^2

10" = 310cm^2

12" = 480cm^2

15" = 810cm^2

so, 4-8's or 3-10's will be as close as your going to get. 3 8's would only be 72% of 1-15 where 4-8's would be 96% of 1-15, 3-10's would be 114% of 1-15

If we were trying to find the exact cone area of the radiating surface, I would agree with you. There are more accurate ways, if for some reason, you need to know EXACTLY. But the point of this thread is comparing cone area between subs, so Pi®^2 will work fine.
Where Pi®^2 falls short is it includes 100% of the surround. Your numbers are likely derived from the 2/3 of surround is included method, but its 'averaged' so who knows what size surround it was factoring. Thus your method is a guess to get close, just like Pi®^2 is. If we/he require exact numbers, look up the cone area for each speaker being compared, as most manufacturers list this spec. Otherwise, if we are just trying to get a rough idea, using Pi®^2 will be fine.

And lets face it, when comparing cone area between systems, if the cone area difference comes out so slight that you need exact cone area figures for each driver being compared to conclude which has more, then other factors like excursion, power input, enclosure design, installation and environment will all play MUCH larger roles in over all output differences.

 

[snb778 5,000+ posts]

3x 10's to a 18"

 

[SomeGuyDude 10+ year member]

pi*r^2 for cone area, so...

x*pi*r(a)^2 = y*pi*r(b)^2 where x and y are given quantities, and r(a) and r(b) are the radii of the given subwoofers. Drop the pi since it's a common term and you can start pulling stuff apart.

x*r(a)^2 = y*r(b)^2

Now let's say you want to know how many 8's are needed for equivalent 15's.

x*r(a)^2 = y*r(b)^2

x*4^2 = y*7.5^2

16x = 56.25y

3.51x = y

7.02x = 2y

So you'd need 7 8's to equal about 2 15's.