Silly Idea

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So I'm building these sealed MTM speakers. The F3 point (as modelled, at least) is around 80Hz. They will be biamped, and it looks like the midranges are going to have a significant amount of additional power over the tweeters (almost an order of magnitude, in Wattage).

Let's assume that power wasn't the problem here. Would it be useful to try to electronically equalize the low-end of the speaker to push the F3 point to about 40Hz, claiming another octave? Or do midranges tend not to have nearly enough excursion? With two midranges, the excursion would be half for a given SPL, correct?

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Joined 2001

The bassmids together must move about 40 cubic inches of air to achieve an SPL of 112 dB at 40 Hz. Cut that by 3/4-down to 10 cubic inches-to achieve 112 dB at 80 Hz. That is both both bassmids together.

To calculate what your bass mids are capable of, multiply the cone area of both mids by their excursion. The excursion should be calculated one way-that is, from center to peak. Peak-to peak excursion will be twice as great, and should therefore be divided by 2 for this calculation.

If your measurements are in metric, then 40 cu.inches equals 625 cubic centimenters.

For every 6 dB less SPL you are willing to accept, cut the amount of air required by half. So the bassmids must move about 20 cu inches of air to generate an SPL of 40 Hz at an SPL of 106.

[Edited by kelticwizard on 12-04-2001 at 11:19 PM]
I guess I'm more interested in where you got those numbers. I assume these numbers are derived from the intrinsic properties of air, but I guess I'm not familiar enough with them.

This is probably not a very good idea for these particular drivers, because WinISD says that the Xmax is quite low.

Thaks, Won
diyAudio Moderator Emeritus
Joined 2001

The numbers are from an article on closed box spekers by Richard Small, (of Thiele-Small fame). Clearly the most respected of sources.

The way I keep track of these things is to remember: 112 dB SPL at 100 Hz. requires about 6 cu. inches of air to be moved.

For every half octave lower at 112 dB you wish to go, double the amount of air. So 70 Hz, (half octave lower) @ 112 dB requires 12 cu. inches.

For every 6 dB increase in SPL, double the volume. So 100 Hz. @ 118 dB requires 12 cu. inches.

To go higher in frequency or lower in SPL, simply reverse the process.

I find it a welcome guide to calculate how much bass you can get from a speaker.

These numbers are for closed box systems only. Ported boxes are a completley different kettle of fish.

[Edited by kelticwizard on 12-05-2001 at 06:51 PM]
I remember an old article by David Weems....

About electronically assisted bass. Sure you can do it but you have to build a correspondingly larger enclosure. It is easier and safer to do with a bass reflex enclosure because the low bass output comes from the port anyway.
This is true. :) I also plan on putting together a subwoofer at some point (lower priority because all of my SACDs are classical/jazz), but I guess I ended up being interested in the relationship between excursion, frequency and sound pressure.

I realized that my idea was basically silly even before I posted, hence the thread title/disclaimer.

Thanks to everyone...Won
diyAudio Moderator Emeritus
Joined 2001
Typo Correction

Oops, had to change my post above.

I changed it to say that for a sound output of 118 dB @ 100 Hz, 12 cu inches of air had to be moved. Before, it said 108 dB @ 100 Hz required 12 cu inches of air to be moved.

Sorry for any inconvenience that caused.

[Edited by kelticwizard on 12-05-2001 at 06:53 PM]
Walker has a very good point. I spent quite a while designing carefully ported boxes for some small drivers to get down to ~ 45Hz for home theatre satellites. Then I added a vented shiva subwoofer and crossed way above that point. It was a waste of time, sealed speakers would have been smaller, easier and sound better. I ended up building sealed boxes for the speakers later anyways.
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