Woofer Speed

[cx-7heaven 7]

Rather than start a new thread, I'll post in this one as the title fits. Got into a discussion on another forum about the actual speed in MPH of a woofer cone at a given frequency. I dont know enough of the math to really figure it out, and I'm sure some of you do. What the other person said was that using an example woofer with 2" peak to peak excursion, the cone would move a total of 4 inches for any given cycle. Multiply that by number of cycles.. in this example 30. Multiply that by 60, and that by 60 again for inches traveled in an hour, divide that by 12 and that by 5280.

So:

4x30x60x60/12/5280=6.82 miles per hour at 30 hz.

My problem is that that seems really slow, I think there is another factor involved which isnt being taken into account. Hopefully one of you folks can help me out.

[aznboi3644 2]

A sub with 2" peak to peak excursion would move a total of 8" in one cycle at peak excursion...not 4".

30hz

8" per cycle...20' per second...blah blah blah comes out to 13.6mph @ 30Hz.

Edited March 23, 2009 by aznboi3644

[BKOLFO4 29]

A sub with 2" peak to peak excursion would move a total of 8" in one cycle at peak excursion...not 4".

Maybe I missed something. . .explain please.

[mpimm 0]

A sub with 2" peak to peak excursion would move a total of 8" in one cycle at peak excursion...not 4".

30hz

8" per cycle...20' per second...blah blah blah comes out to 13.6mph @ 30Hz.

That is average speed over the cycle. I think what he is curious about is instantaneous velocity which is a maximum when displacement is 0 and can be solved taking the derivative of the displacement function.

@30Hz

s(t)=2"sin(at) displacement=0:t=0 ax=0 max displacement: at=pi/2 t=(1/120)s a=pi*60/s,

ds/dt=v(t)=2"cos(at)*a

vmax=2"*pi*60/s~=377"/s or 21.4mph

... I think...

[cx-7heaven 7]

I got a headache looking at that formula, but 21 mph is starting to sound much more realistic.

[95Honda 486]

Another quick analogy I like to use to help people understand Mms is like this-

You take driver A) with a Mms of 100gr and driver B) (an identical driver) with a Mms of 100gr.

Now, put these two together face-to-face isobarik and you have a combined Mms of 200gr (+ whatever the air weighs in between them)

Now, you have doubled Mms, so transients should suffer.... When in fact they don't.... They don't because you now have twice the motor force for twice the Mms. Everything equals out, even though you have more Mms. This goes to show you can still have Mms that is larger if you counter it with motor force, all while saving "transient response"... Because honestly, isobarik alignments are know for alot of things, but degraded transient performance over that of a single driver isn't one of them....

But really, this isn't transient response so much as efficiency in relation to Mms.... No matter how heavy a cone is it will still move back and forth as fast as the AC signal, it will just move less given more mass with identical motor force...

I think Le has a lot to do with transient performance and Mms has a lot to do with efficiency, within reasonable limits, of course...

I also think heavy cones with lots of inductance render a driver with poor efficiency and poor transient behavior... Even if it has 5" of xmax...

[aznboi3644 2]

Isobaric loading also cancels out driver linearity problems.

[DevilDriver 28]

x = x0 + vt

x = 4"

x0 = 0"

t = 1 second/30 cycles = 1s/60

Solve for v. I divide the time by 60 as there are two 4" movements per cycle; as an alternative, you could set x to 8" and leave t as 1s/30 cycles.

As I've mentioned several times, Le is the dominant aspect of transient response, while Mms (and other factors) play a lesser role. Note that there is a relationship between the time domain and the frequency domain.