The free air resonance parameter (Fs) on a speaker driver is the frequency at which the driver resonates most strongly when mounted in a free-air environment (i.e., not in a cabinet). This means that at Fs, the driver's cone will vibrate with the greatest amplitude, potentially leading to distortion and reduced sound quality if not properly controlled.
Here are some key points to understand about Fs:
* Lower Fs: Generally, a lower Fs indicates a deeper bass response. However, it also means that the driver may require a larger or more complex enclosure to achieve optimal performance.
* Higher Fs: A higher Fs typically results in a more controlled bass response, but it may also limit the driver's ability to reproduce deep frequencies.
Here are some key points to understand about Fs:
* Lower Fs: Generally, a lower Fs indicates a deeper bass response. However, it also means that the driver may require a larger or more complex enclosure to achieve optimal performance.
* Higher Fs: A higher Fs typically results in a more controlled bass response, but it may also limit the driver's ability to reproduce deep frequencies.
I know what Fs is. I just don't know what it is for this driver. I'm guessing it's around 40-42hz, this the frequency where the cone is most active.
If you would like to measure your driver’s resonant frequency (Fs), connect the “+” output of a frequency generator/amplifier output directly to the driver “+”, place a 10 ohm resistor in series with the driver “-“ . Connect an AC voltmeter across the resistor leads. Adjust the frequency up and down until the voltage across the resistor reaches it’s minimum. The voltage minima is the resonant frequency.
The 10" probably has a looser suspension than the 8".
The compliance of a driver's suspension has a significant impact on its behaviour in response to very low frequency input, in free air conditions. High compliance (or low mechanical resistance, which is another way of saying the same thing) tends to result in high unloaded excursion at LF.
(Note, in relation to the thread title, that lots of LF excursion when unloaded does not necessarily imply extended LF frequency response.)
(Note, in relation to the thread title, that lots of LF excursion when unloaded does not necessarily imply extended LF frequency response.)
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I'm not that serious about speaker building. But I've learned a lot asking questions on this forum.
Are you serious enough to buy a few pounds worth of resistors and solder to measure via your computer and REW?
No. Electronics really isn't my bag. Science isn't really a big interest for me. The are a couple of OGs on here, people who can look at a driver, feel the mechanical resistance, observe the x-max, tap the diaphragm and estimate the resonant frequency, consider the magnet weight, and decide driver X is likely preferable to driver Y for said task.
I don't know - you asked a question in your initial post that can only be answered by science.
Just to add to my previous post, I did a quick sim of a couple of drivers, a Scanspeak 10" woofer, and a Dayton 8" sub. The Scanspeak has high compliance, for a 10" - Cms is 0.95mm/N. The Dayton has lower compliance - Cms is 0.29mm/N. So in theory if you apply a driving force equivalent to 1 Newton, the Scanspeak's cone will move 0.95 millimetres, the Dayton's cone only 0.29mm.
How does that work out in practice? The sim suggests that if you apply a 2.83V signal at 10 Hz, with the drivers in free air, the Scanspeak will have a peak excursion of 4.8mm, where the Dayton's peak excursion would only be about 2.4mm. Both drivers have very similar impedance, so are drawing the same power. The difference in compliance is the main factor, I reckon.
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