if a subwoofer speaker need 2ft cubic, can i put 4 subwoofer speaker in the same volume ?
Hi,
Yes you can: In HR the configuration would be 2P,2s or 2S,2p...
b🙂
For the sake of the OP, do you mean if you run two sets isobaric? If so, do you still have to add a bit of volume to account for the lost space when running them that way, as in stacked? Or are you suggesting face to face with the mags sticking out?
Hi Cal,
I admit my answer was cryptic, Here is more 😀 ..
One driver(one cone area facing the listener) in one box = two clam-shelled drivers(two cone areas(Magnets) facing the listener) using the same volume as for one driver.
I estimate that the additional volume occupied by the second isobaric speakers(good sub drivers for home use) would be doubled but compared to the total available volume of 2 cu.ft this would be of a minor issue if we aren't talking about huge Car speakers 😱 with a rear side zero clearance to the enclosure walls or magnets.
b🙂
I admit my answer was cryptic, Here is more 😀 ..
One driver(one cone area facing the listener) in one box = two clam-shelled drivers(two cone areas(Magnets) facing the listener) using the same volume as for one driver.
I estimate that the additional volume occupied by the second isobaric speakers(good sub drivers for home use) would be doubled but compared to the total available volume of 2 cu.ft this would be of a minor issue if we aren't talking about huge Car speakers 😱 with a rear side zero clearance to the enclosure walls or magnets.
b🙂
thanks, but if i put 4 speakers in a box of 2 cubic feat . each speaker works like its have 2 cu.ft
thanks, but if i put 4 speakers in a box of 2 cubic feat . each speaker works like its have 2 cu.ft
Hi,
If one speaker-driver is requiring 2 cu.ft then if two of these drivers is put in a clamshell system configuration only 1 cu.ft is required , thus 2 cu.ft could accommodate 4 drivers, i.e. = 2 isobaric configured driver systems.
b🙂
not is a isobaric box is a bass reflex or sealed
It could be either...or..
b 🙂
Again, just to remove the confusion, no you can't just add 4 drivers to a 2 cu ft box, you must run them isobaric or you have to increase the box by 2 cu ft every time you add another driver.
thanks, but if i put 4 speakers in a box of 2 cubic feat . each speaker works like its have 2 cu.ft
No. Each driver would work like it has only .5 cu.ft. 2/4 =.5
edit: Cal, +1
Edward,
Ignore all this isobaric "stuff." I doubt you were asking the question with that driver arrangement being considered.
I believe what you are asking is if a box of a certain volume will "give" each driver loaded into it, that "volume" to work with. The answer is No, the volume of the box will be "divided" amongst the drivers. (This is a very crude explanation).
Technically you can cram as many drivers as you want on a box, however, a different number of drivers within the same box will produce a different response.
Eric
Ignore all this isobaric "stuff." I doubt you were asking the question with that driver arrangement being considered.
I believe what you are asking is if a box of a certain volume will "give" each driver loaded into it, that "volume" to work with. The answer is No, the volume of the box will be "divided" amongst the drivers. (This is a very crude explanation).
Technically you can cram as many drivers as you want on a box, however, a different number of drivers within the same box will produce a different response.
Eric
Eric, you ask him to ignore the good advice about how to put 4 drivers into a box designed for 1 and then tell go on to tell him what we have already said about normal alignments. What gives?
Eric, you ask him to ignore the good advice about how to put 4 drivers into a box designed for 1 and then tell go on to tell him what we have already said about normal alignments. What gives?
Agree with Cal,
And IMO: If scrutinizing the suggested nonsense: ' the box will be "divided" ' ..or.. 'would work like it has only':
The last two posts above the one of yours, have no foundation in the reality whatsoever, if compared to any trustworthy schematic concepts that uses analog lumped electrical circuits like fundamental TS or derived parametrized data explaining the working physical or functionals of a speaker:
Instead, they are layman concepts that do nothing but further: Is a disorientation what it's all about.
Show me a schema or any mathematics that divides a shared volume into different partitions in order to solve the load situation(acoustic too) exposed to each driver that really would work and gives a good picture for a newbie to understand and is a good reference for knowledge worth saving for future DIYaudio projects.
b 🙂
Hi Cal,
I read through the thread and felt that the issue was being over-complicated. Post #6 proves that the answers being given were overcomplicated. I tried to help clarify. Now look what's happening... Further over complication that will do the OP absolutely no good at all. Good job bjorno.
This guy needs baby steps and you're all throwing him off a cliff and expecting wings to sprout, then giving me a hard time for trying to help clarify. This is very frustrating for me.
Regards,
Eric
I read through the thread and felt that the issue was being over-complicated. Post #6 proves that the answers being given were overcomplicated. I tried to help clarify. Now look what's happening... Further over complication that will do the OP absolutely no good at all. Good job bjorno.
This guy needs baby steps and you're all throwing him off a cliff and expecting wings to sprout, then giving me a hard time for trying to help clarify. This is very frustrating for me.
Regards,
Eric
I'm a bit perplexed by B's comments.
Since Qtc = sqrt (( Vas / Vb ) + 1 ) x Qts, the ratio of Vas to Vb must remain the same if a specified Qtc is to be maintained.
Since Qtc = sqrt (( Vas / Vb ) + 1 ) x Qts, the ratio of Vas to Vb must remain the same if a specified Qtc is to be maintained.
cal weldon, can you explain me the reason because i cant ?
OK, we'll keep this understandable.
Each woofer needs a certain amount of air to 'breathe' to do it's job. Let's say you have a woofer that needs 2 cu.ft to do the job. If you put 3 more woofers in there, the air has to be divided among them so in essence they are only getting .5 cu.ft of air each so they struggle to breathe as they are all working 'against' each other in a 'normal' configuration.
Isobaric loading allows you to couple the drivers. This means that you are basically halving the cone area but using two motors for each of those cones and working them together. Because they are working together, they only need 1/2 the original cabinet size per 'set' so now you can use 2 'sets' together in that original 2 cu.ft box.
The drawback is, unlike a normal installation where you get to add dB's as you increase the # of woofers, in this case you don't, 4 woofers will give you the same volume per watt as 1 woofer.
The advantage of it is you will have a higher power rating.
Try this from Wiki:
Isobaric speakers - Wikipedia, the free encyclopedia
Agreed but isobaric is not really complicated.Further over complication that will do the OP absolutely no good at all.
I am hoping my above post helps because it sounds like he wants to use 4 woofers in one 2 cu.ft box.This guy needs baby steps and you're all throwing him off a cliff
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