What speaker specs for a free air install (aka car door)? and other suggestions

MITMechE · 11th May 2002, 02:42 PM

Hello folks, I'm new to this forum you have a nice place here.

What T/S specs should I look for ballpark for a free air install of a woofer? I'm planning on fitting an 8" speaker into my door and putting a ~4" mid in a small kickpanel. I searched and only saw one suggestion for a QTS of around .7, is this the best value? What about Vas?

The 4" Mid needs to have a very small sealed volume. So should I look for T/S specs of small Vas and small QtS? Anyone have any good recommendations for a relatively cheap (<$30) 4 ohm mid that might work in my install?

Thanks guys.

kelticwizard · 11th May 2002, 04:42 PM

I just cut-and-pasted part of an earlier post that deals with what happens with speakers of a certain Fs and Qts, when put into a sealed enclosure. As the enclosure become several times the speaker's Vas, it resembles a free air mounting.

If you plan to mount a speaker with a Vas so small that the car door volume is twice the volume of the Vas or more, then .7 is the way to go.

"As for closed box systems, you want a Qtc, (final Q of the closed box/driver combination), of between .7 and 1.0. Purists like a final Qtc of .5, but your output is 6 dB down at resonance in that alignment.

Fc is the resonance frequency of the closed box/driver combination.

When a driver is placed into a closed box, BOTH it's Qts and Fs are raised according to the same formula. For the raise in Q, the following formula is used:

Qtc = the square root of [(Vas/Vb) + 1] times Qts.

And applying the same formula to the resonance frequency:

Fc = the square root of [(Vas/Vb) + 1] times Fs.

So, a driver with a Qts of .5, if placed into a box that is one third it's Vas, will yield a driver/box combination that has a Qtc of 1.0 and an Fc that is twice the drivers Fs.

Any number of programs will give you these answers, but I thought I would give you some reasons as to why they arrive at these numbers. You can make your design decisions based on these principles"

Hope this helps. Good luck!

BAM · 11th May 2002, 09:52 PM

This is an 8" subwoofer that will work in a free-air system, made by JL audio. They also have free-air 10's and 12's.
Parts Express sells some 4" PYLE midbasses and matching tweeters.

11th May 2002, 02:42 PM	#1
MITMechE diyAudio Member Join Date: May 2002 Location: Cambridge, MA	What speaker specs for a free air install (aka car door)? and other suggestions Hello folks, I'm new to this forum you have a nice place here. What T/S specs should I look for ballpark for a free air install of a woofer? I'm planning on fitting an 8" speaker into my door and putting a ~4" mid in a small kickpanel. I searched and only saw one suggestion for a QTS of around .7, is this the best value? What about Vas? The 4" Mid needs to have a very small sealed volume. So should I look for T/S specs of small Vas and small QtS? Anyone have any good recommendations for a relatively cheap (<$30) 4 ohm mid that might work in my install? Thanks guys.

11th May 2002, 09:52 PM	#3
BAM diyAudio Member Join Date: Sep 2001 Location: Carmel, IN; West Lafayette, IN	JL Audio 8IB4 This is an 8" subwoofer that will work in a free-air system, made by JL audio. They also have free-air 10's and 12's. Parts Express sells some 4" PYLE midbasses and matching tweeters.

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11th May 2002, 04:42 PM	#2
kelticwizard diyAudio Member Join Date: Sep 2001 Location: Connecticut, The Nutmeg State	I just cut-and-pasted part of an earlier post that deals with what happens with speakers of a certain Fs and Qts, when put into a sealed enclosure. As the enclosure become several times the speaker's Vas, it resembles a free air mounting. If you plan to mount a speaker with a Vas so small that the car door volume is twice the volume of the Vas or more, then .7 is the way to go. "As for closed box systems, you want a Qtc, (final Q of the closed box/driver combination), of between .7 and 1.0. Purists like a final Qtc of .5, but your output is 6 dB down at resonance in that alignment. Fc is the resonance frequency of the closed box/driver combination. When a driver is placed into a closed box, BOTH it's Qts and Fs are raised according to the same formula. For the raise in Q, the following formula is used: Qtc = the square root of [(Vas/Vb) + 1] times Qts. And applying the same formula to the resonance frequency: Fc = the square root of [(Vas/Vb) + 1] times Fs. So, a driver with a Qts of .5, if placed into a box that is one third it's Vas, will yield a driver/box combination that has a Qtc of 1.0 and an Fc that is twice the drivers Fs. Any number of programs will give you these answers, but I thought I would give you some reasons as to why they arrive at these numbers. You can make your design decisions based on these principles" Hope this helps. Good luck!