Discussion arising from Geddes loudspeaker

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Joined 2007
dlr said:


It was not I who started it. Gedlee pointed out the error in his post #361


I actually pointed out Earl's error in his post #330. That's what started this.

Gedlee:
My bet is on the direct excitation of the panels by the loudspeaker as opposed to the coupling through the air. Air coupling falls off as 1/f, but direct coupling of the vibrations is almost constant.
 
MJL21193 said:



I actually pointed out Earl's error in his post #330. That's what started this.

"Gedlee:
My bet is on the direct excitation of the panels by the loudspeaker as opposed to the coupling through the air. Air coupling falls off as 1/f, but direct coupling of the vibrations is almost constant."


And the statement is just as correct now as when I first made it. None of this nonsense is going to change that.
 
dlr said:


...
The opposing force will be constant with frequency if the original applied force is constant with frequency. Beyond that it's trivial.

...
Dave
The opposing force cannot be constant with frequency because of the effective mass changes at higher frequencies. It may be constant over a limited bandwidth, but certainly not the whole audio spectrum.
 
frugal-phile™
Joined 2001
Paid Member
dlr said:
"The further the cone travels, the more opposing force, therefore more kinetic energy to deal with.".

the important part there is more kinetic energy?

There will be no transfer of energy without force

In some cases down at the atomic level.

It takes force to move an object, it's the force that matters.

But you can apply a force, not move the object, and no work is done. When the force moves the object you have work = energy

The force is needed to do the work, but the resulting work (energy) is the metric which we need to consider -- how much energy is injected into the panel.

fact that total transferred energy drops with increasing frequency is not relevant,

It is VERY relevant.

dave
 
Disabled Account
Joined 2007
Let's try this again.

The drivers cone is pushed forward by the voice coil. Forward motion of the cone acts on the driver frame to push this back. Newton's third law of motion. Clear? Picture recoil of a rifle - bullet is fired forward, gun goes backwards.

The further the cone goes forward, the more the driver frame will tend to go back. Since you need longer excursion at lower frequencies to produce the same spl, this results in further cone travel and more rearward driver motion in reaction to it.
Obvious to me that at the higher frequencies, where large cone excursion is not needed to achieve higher spl levels, there will be less reactionary motion.

The last time I used words like "force" and "energy" and these just confused a few, so I carefully avoided them here.
 
MJL21193 said:
Let's try this again.
The further the cone goes forward, the more the driver frame will tend to go back. Since you need longer excursion at lower frequencies to produce the same spl, this results in further cone travel and more rearward driver motion in reaction to it.
Obvious to me that at the higher frequencies, where large cone excursion is not needed to achieve higher spl levels, there will be less reactionary motion.
There is either misunderstanding of the physics here or more likely confused wording.

The displacement of the driver (frame) depends on the force and the stiffness or mass of the box*. At a frequency where the box is rigid and, because it is massive, barely moves, there is little energy transferred to it (force times displacement). At a frequency where it is softer (at a resonance) the displacement is more, and more energy is transferred for the same force.

You can feel that happening if you apply a swept sine and feel the baffle. Most energy is transferred to your fingers at resonances of the baffle (or of the speaker rocking on its support).

*In simplified form: the force (F) leads to a compression (x) of a massive box acording to the effective stiffness (k) as a function of frequency (think of x = F/k) , and to an acceleration (a=(2*pi*f)^2) of a stiff box according to its mass (a=F/m). In many cases the motion due to the flexing of the box will exceed that due to recoil at box resonances.

Pick your own force spectrum to put in (either real music or the models already mentioned).

Ken
 
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