| Septimus |
I just need to know the volume needed to obtain an optimally flat response -in car- for a sealed enclosure. that means calculated with cabin gain. I have a protege wagon if cabin size can be taken into account.
I just googled average interior volume of a car and hit wikiepedia.
http://en.wikipedia.org/wiki/Vehicle_size_class
Scroll down to USA regulations, small station wagon says under 130, over 120.
Anyway. I have a mac and am too lazy to drag the pc laptop out of the closet and find some software for it.
Here are the specs on my sub. I will be feeding it 300 watts.
Here are the technical specs;
Dual 4 ohm
Revc = 7.8 Ohm
Fo = 22.6 Hz
Sd = 350.0 cm^2
Krm = 2.1538m Ohm
Erm = 1.085
Kxm = 193.204m H
Exm = 0.588
Vas = 71.3 Ltr
Cms = 409.87u M/N
Mmd = 129.248m Kg
Mms = 133.013 g
BL = 19.032 T.M
Qms = 3.290
Qes = 0.37
Qts = 0.333
No = 0.22%
SPLo = 85.36 dB
Thanks in advance! |
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| Septimus |
| I have a mac and don't have virtual PC, so I can't use any of the free software online. |
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| noodle_snacks |
| try taking a look, thats an online version of winisd |
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| bibster |
Hey, that's some handy page!
Now I can finallt get rid of this ol'PC. WinISD was like the last reason to keep this PC up and running!!
(Other audio/speaker stuff related MAc software hints are welcome, though!)
Thanks, Paul |
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| richie00boy |
Cabin gain is determined by the lengths of space inside the car, not volume. Give us the length and width of the internal cabin space and we can make a reasonable guess as to the cabin gain frequency.
Then you set F3 of the box to match this, with Qts dependent on sound preferences and length/width ratio of the cabin. |
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| Septimus |
| It is a mazda protega wagon. Full length is probably 10-11ft, width is 5ft. |
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| richie00boy |
| That will give you 51-57Hz to aim for with Qts in the range 0.65-0.8 |
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