Smallest sealed box w/ 100% stuffing for RSS265HF-4 ohm 10" w/ Q of .70?

**costanza** · **11-05-09, 03:05 PM**

Taking the displacement of the driver into account. Thanks

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**Mike P.** · **11-05-09, 06:53 PM**

The RSS265HF-4 has a Q of .7 at 3 cu.ft. Assuming 100% stuffing equals a 25% increase, then the box would be 2.4 cu.ft. Add whatever the displacement of the sub is to that and you have the total internal volume.

**costanza** · **11-05-09, 08:43 PM**

Quote:

Mike P. wrote:

The RSS265HF-4 has a Q of .7 at 3 cu.ft. Assuming 100% stuffing equals a 25% increase, then the box would be 2.4 cu.ft. Add whatever the displacement of the sub is to that and you have the total internal volume.

Hmm? Are you sure you modeled the 10"? I have never used WinISD before, but I entered the driver data into it and it says Q of .703 = 1.098 cu ft. sealed. Now, subract 25% of that volume for 100% stuffing = .82 cu ft. (+ air displaced by driver itself). Is this wrong? How do I determine how much air the driver is displacing? Thanks for all the help!

**Mike P.** · **11-05-09, 09:05 PM**

Here is the WinISD file for the sub. Check these parameters against the ones you entered into the program.

Dayton 10_ Reference HF (RSS265HF-4).wdr.

**costanza** · **11-05-09, 09:45 PM**

That is the file I saved to the drivers folder.
I chose that driver, in a sealed enclosure and I chose the .7 Q selection.
When I clicked on the box tab, it read 1.098 cu ft.
75% of that (for 100% stuffing) is .82 cu ft.
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**costanza** · **11-06-09, 01:13 AM**

It seems that other sites also recommend 1 cu ft sealed for this driver too... >confused<
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**Mike P.** · **11-06-09, 04:56 AM**

There's a problem somewhere. Not sure where though. Hopefully someone else can verify which is correct.

Quote:

It seems that other sites also recommend 1 cu ft sealed for this driver too

Are they just recommending 1 cu. ft. sealed or are they saying 1.cu. ft. is a Q of .7?

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**costanza** · **11-06-09, 10:13 AM**

The driver files we're using are different. Yours has ohm's as being 6, vs. 4 and a couple other diffrences.

**costanza** · **11-06-09, 10:27 AM**

The box I have already built is roughly 1.6 cu ft.
I subtract .4 for an internal amp enclosure and subtract .10 for an estimate of the driver's displacement, which gives me just at 1 cu ft. +/- 10% (no stuffing). This sound good?

**costanza** · **11-06-09, 11:35 AM**

How weird is this?! My box's internal volume just happens to be 1.64 cu ft (14.5 x 13.5 x 14.5) with the amp enclosure removed. Winisd's recommendation for vented enclosure for this driver is 1.43 cu ft (bb4/sbb4) w/ 1 vent 4" x 31", which itself equals .22 cu ft. Subtract the volume of the port's tubing from my box's volume, and you have 1.42 cu ft.!!!!!

**Mike P.** · **11-06-09, 01:54 PM**

Quote:

The driver files we're using are different. Yours has ohm's as being 6, vs. 4 and a couple other diffrences

Re is listed as 4.4 ohms in the file as stated from the parameters. Any small differences are from the auto calculate feature in the program which won't affect the modeling. Upload the file you are using and I'll try and find the problem.

**costanza** · **11-06-09, 02:13 PM**

http://www.hometheatershack.com/foru...ference-10.wdr

**Mike P.** · **11-06-09, 09:43 PM**

Your file is correct. Some how mine got corrupted. The parameters in mine were way out, not sure how that happened.

**evilskillit** · **11-07-09, 11:13 AM**

Yeah I checked the file I submitted vs the PDF and it looks good. 1 cu ft is good for sealed. 1.5 is good but small for vented, one problem with venting these small drivers is for a very low tune in a smallish box the ports get real long real quick. A 4" port needs to be liked 35-40" for a low tune in a 1.5 cu ft box. Try fitting a port that long in a box that small... ? Anyways either way good luck with your build, let us know where you end up.

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