Forgive the long rambling nature of this post...
The TLDR version; is a 7/8" deep port in a 5/8" thick enclosure wall really aperiodic or ported?
-If its the former, how much of an affect will 25% less volume really have
-if its the later, how much airflow can an odd shape enclosure be choked without affecting response (ie bottom picture)
I'm forging ahead fitting a 3 way cabinet speaker into a shelving unit I made.
Its an old 1990 model. Internal volume is 2.1³ ft. The tweeter MW are encased from behind in and have a high XO, all that volume is used by the 12" woofer (which runs full range).
It's spec'ed as ported, frequency response 45hz-20khz but the port is 1.8" diameter, 0.9" deep (very slightly thicker than the wood).
The shelving unit can fit 1.35³ ft enclosures. Stuffing fiberglass in theory makes the relative size 75% of the original size. Without t/s parameters or modeling is there any predictions on how the reposnse would chance?
If I lose some of the bass extension but the response above 110hz is mostly unchanged I can live with that. (I think that has more to do if the original really is a ported enclosure or aperiodic.
With that said, I can increase the volume using odd size extensions to the enclosure, but I think my ideas would create more problems even if the total volume is identical, right?
The TLDR version; is a 7/8" deep port in a 5/8" thick enclosure wall really aperiodic or ported?
-If its the former, how much of an affect will 25% less volume really have
-if its the later, how much airflow can an odd shape enclosure be choked without affecting response (ie bottom picture)
I'm forging ahead fitting a 3 way cabinet speaker into a shelving unit I made.
Its an old 1990 model. Internal volume is 2.1³ ft. The tweeter MW are encased from behind in and have a high XO, all that volume is used by the 12" woofer (which runs full range).
It's spec'ed as ported, frequency response 45hz-20khz but the port is 1.8" diameter, 0.9" deep (very slightly thicker than the wood).
The shelving unit can fit 1.35³ ft enclosures. Stuffing fiberglass in theory makes the relative size 75% of the original size. Without t/s parameters or modeling is there any predictions on how the reposnse would chance?
If I lose some of the bass extension but the response above 110hz is mostly unchanged I can live with that. (I think that has more to do if the original really is a ported enclosure or aperiodic.
With that said, I can increase the volume using odd size extensions to the enclosure, but I think my ideas would create more problems even if the total volume is identical, right?
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Let's see if I've got this right - you are intending to house a 12" woofer in a 1.35 ft³ enclosure instead of in its original 2.1 ft³ enclosure?
If so, the original port dimensions would not suit the smaller enclosure.
Densely stuffing an enclosure with absorbent material will lower the system Q and can be used to flatten a bass hump in the response curve.
P.S. Aperiodic loading involves a resistive port - typically a series of 1/2" diameter holes backed with soft cloth to resist the flow of air - not the same as an open port which allows free flow of air.
P.P.S. If you you wish to extend the 1.35 ft³ volume, then I would avoid the narrow passageway to the additional chamber shown in your left hand diagram.
If so, the original port dimensions would not suit the smaller enclosure.
Densely stuffing an enclosure with absorbent material will lower the system Q and can be used to flatten a bass hump in the response curve.
P.S. Aperiodic loading involves a resistive port - typically a series of 1/2" diameter holes backed with soft cloth to resist the flow of air - not the same as an open port which allows free flow of air.
P.P.S. If you you wish to extend the 1.35 ft³ volume, then I would avoid the narrow passageway to the additional chamber shown in your left hand diagram.
Changing a ported box to a sealed box usually involves a reduction in volume.
So you may find a 1.35 ft³ enclosure to be satisfactory provided you seal it and experiment with the stuffing density to tailor the bass response.
Alternatively, the 1.35 ft³ enclosure could make use of aperiodic loading to reduce the Q of the bass resonance and extend it downwards in frequency compared to the sealed enclosure. This involves drilling a series of 1/2" holes in the rear of the enclosure and covering them internally with soft, felt like cloth. In this case, you would only apply absorbent to the walls of the enclosure rather than fill it completely. I calculate that 7 rows of 10 holes should be about right for the job, but there is room for experimentation. Space the rows of holes 1.5 to 2 inches apart.
So you may find a 1.35 ft³ enclosure to be satisfactory provided you seal it and experiment with the stuffing density to tailor the bass response.
Alternatively, the 1.35 ft³ enclosure could make use of aperiodic loading to reduce the Q of the bass resonance and extend it downwards in frequency compared to the sealed enclosure. This involves drilling a series of 1/2" holes in the rear of the enclosure and covering them internally with soft, felt like cloth. In this case, you would only apply absorbent to the walls of the enclosure rather than fill it completely. I calculate that 7 rows of 10 holes should be about right for the job, but there is room for experimentation. Space the rows of holes 1.5 to 2 inches apart.
Without getting too weird with the shape (back is a little taller than the front) I got 1.53 cu ft.
The original box 2.06cu ft, Tunning frequency 40hz. Fs is 138hz.
If I adjust the port to tune it to 50hz+ and 1.53cu ft. I have a pretty good idea of what happens bellow Fs (it's covered in detail in 100's subwoofer guides)
This is a full range woofer high end rolls off around 9khz.
How much will smaller enclosure, tuned higher affect the response graph from Fs and up (138hz-9khz).
Am I going to lose sensitivity and add spikes/dips in the response?
Or should I keep it at 1.35 ft³ with the aperiodic design you suggested?
the external subwoofer is handling <120hz, so the woofers low end response is getting mostly filtered... I don't want to get bogged trying to make a cross over for drivers with non-published specs like this one.
The original box 2.06cu ft, Tunning frequency 40hz. Fs is 138hz.
If I adjust the port to tune it to 50hz+ and 1.53cu ft. I have a pretty good idea of what happens bellow Fs (it's covered in detail in 100's subwoofer guides)
This is a full range woofer high end rolls off around 9khz.
How much will smaller enclosure, tuned higher affect the response graph from Fs and up (138hz-9khz).
Am I going to lose sensitivity and add spikes/dips in the response?
Or should I keep it at 1.35 ft³ with the aperiodic design you suggested?
the external subwoofer is handling <120hz, so the woofers low end response is getting mostly filtered... I don't want to get bogged trying to make a cross over for drivers with non-published specs like this one.
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I'm sorry, but I can't predict the effect on the response of tuning the smaller enclosure to 50 Hz - that would call for an expert like GM!
I can predict that, if the 12" driver is running full range, the aperiodic enclosure will provide an upper bass performance that is resonant free, resulting in good transient response and an 'open' sound. This is especially beneficial in providing clarity to deep male voices.
The external subwoofer wll take care of the lower bass frequencies.
You could experiment in three stages. Stop at the stage where you think the sound balance is satisfactory.
I can predict that, if the 12" driver is running full range, the aperiodic enclosure will provide an upper bass performance that is resonant free, resulting in good transient response and an 'open' sound. This is especially beneficial in providing clarity to deep male voices.
The external subwoofer wll take care of the lower bass frequencies.
You could experiment in three stages. Stop at the stage where you think the sound balance is satisfactory.
- Start with a sealed enclosure and experiment with the density of absorbent filling.
- Install a port as per your calculations and attach absorbent material only to the internal walls.
- Block the port and convert the enclosure to aperiodic.
This sounds like a plan. I'm going to have go with 8 x 1.5" holes.
The Back is 11.75X11.75 Id only fit 25 maybe 36 1/2" holes if I use the entire back. Otherwise, it's a plan.
Follow up question you might know (and probably doesnt matter)
Should I stable paper backed fiberglass insulation to the inside wall as the "stuffing" or glue polyfill batting?
Am I better off using laminate coated mdf on the inside of the enclosure or bare mdf? (I have double sided. single sided, and all bare sheets already cut to size).
Unrelated side note, I've been frustrating myself cutting different shape enclosures that I made pvc board stencils to make 1/2" and 3/4" panel enclosures that fit standard cubbies 12.75"x12.75"x (upto 18" depth)
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I would not use 1.5 inch holes as they are unlikely to provide sufficient resistance to air flow.
I'll come back to you on that after studying the dimensions of the rear panel.
P.S. Fs is the free air resonant frequency of the loudspeaker cone, measured when the driver is suspended vertically in the air.
I would use the heaviest (densest) mdf sheets, It also could be argued that surface laminated both sides gives greater strength.
P.P.S. I would favour gluing the polyfill.
Edited after your edit!
I'll come back to you on that after studying the dimensions of the rear panel.
P.S. Fs is the free air resonant frequency of the loudspeaker cone, measured when the driver is suspended vertically in the air.
I would use the heaviest (densest) mdf sheets, It also could be argued that surface laminated both sides gives greater strength.
P.P.S. I would favour gluing the polyfill.
Edited after your edit!
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Let's go for 64, 1/2" holes, in an 8 by 8 array like the one shown in the attachment.
N.B. The centres of the holes shall be 3/4" apart.
I reckon that the array will measure 5.75" by 5.75" and fit comfortably on your 11.75" by 11.75" enclosure back.
But don't take my word for it, draw the array out full size on a sheet of paper to confirm my mathematical ramblings!
Attachments
Because 5 holes of 1/2" diameter result in a total area of 1 square inch! 😉
Which makes the maths easy while being small enough to offer sufficient resistance (in combination with the soft cloth backing).
Which makes the maths easy while being small enough to offer sufficient resistance (in combination with the soft cloth backing).
P.S. I'm simply using the rule of thumb that the area of the resistive vent shall be 10 square inches per cubic foot of enclosure volume.
Taking your internal enclosure volume as 1.3 cu ft, gives us 13 square inches of total vent area.
13 square inches requires (13 x 5) = 65 holes.
The number of holes does not need to be exact as there is a lot of leeway in regard to the design of an aperiodic enclosure.
Taking your internal enclosure volume as 1.3 cu ft, gives us 13 square inches of total vent area.
13 square inches requires (13 x 5) = 65 holes.
The number of holes does not need to be exact as there is a lot of leeway in regard to the design of an aperiodic enclosure.
I've just noticed that, in post #5, you re-estimated the internal volume to be 1.53 rather than 1.35 cubic foot. I did not at first see the transposition of the numbers 3 and 5!
Anyhow, you now have the maths and can adjust the number of holes accordingly.
Anyhow, you now have the maths and can adjust the number of holes accordingly.
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I can't have holes on 2 opposite sides for an aperiodic setup, right? Repurpose the port holes instead of making a 2nd
Long winded reason why : the MDF is from office furniture I dragged out of a lawfirm dumpster. Thick Laminate on both sides, 1/4" of normal MDF sandwiching a 1/4" layer of darker much denser/darker fiberboard (HDF maybe?)
I've been trying to find a use for it, too heavy for most projects (I used one piece for the base of a cat tree).
A cube made from Homedepot 3/4" MDF weight 20lbs, this stuff weighs 30lbs, regular drill bits burn, I went through 2 1/8" cobalt bits for the 20 or so screw holes. I dread trying to make that many holes with a 1/2" bits.
Good, the stuff Im using is tough to drill. Im going to use make 21 holes using a 23mm (0.9") diamond hole saw bit-smallest I have.
Long winded reason why : the MDF is from office furniture I dragged out of a lawfirm dumpster. Thick Laminate on both sides, 1/4" of normal MDF sandwiching a 1/4" layer of darker much denser/darker fiberboard (HDF maybe?)
I've been trying to find a use for it, too heavy for most projects (I used one piece for the base of a cat tree).
A cube made from Homedepot 3/4" MDF weight 20lbs, this stuff weighs 30lbs, regular drill bits burn, I went through 2 1/8" cobalt bits for the 20 or so screw holes. I dread trying to make that many holes with a 1/2" bits.
You're correct the first time. I wouldn't be able to use the larger enclosure and have reasonable clearance behind it. Testing 1.35 aperiodic vs 1.53 ported
I'm not quite sure what you mean when you you say repurpose the port holes.
I suppose the aperiodic holes could be split between two sides of the enclosure, for example, between the front and back.
P.S. All of this is highly experimental and I can't guarantee good results. You would be advised to do a mock-up using normal, thin mdf first, before committing yourself to working with the rock-hard stuff!