Cone movement vs. frequency

[Nichol1997]

I am looking for the mathematical relationship between cone movement versus frequency. Lets say if a subwoofer was held in free-air and the amplitude was constant and the frequency was swept from 100 Hertz down, does the cone move more as the frequency is lowered?

[sreten]

Hi,

Yes it does. Excursion therextically would quadruple at 50Hz, and
be times 16 at 25Hz, but box alignments and reflexing change all
that. Use WinISDpro to look at excursion.

/sreten.

[Svante]

Sreten is right (as usual). If the frequency is lowered, the cone moves more. If the box is a closed box, the cone amplitude will be approximately proportional to 1/f² down to the cutoff frequency of the subwoofer. Below that frequency the amplitude of the cone will be approximately constant, but the sound level will decrease, approximately proportional to f².

[AndrewT]

Hi,
I am kind of new to this speaker lark, but that does not sound right.

I imagine a DC of 1Volt will cause the speaker Voice Coil to move to a new position. Let's suppose for a moment it moves 5mm.
Now apply a -1Vdc to the speaker and it will move the other way 5mm( provided the motor is symetrical).
Now apply a very low frequency square wave of 2Vpp and the speaker will move 10mmpp i.e. +-5mm.
Now apply a very low frequency sine wave of 2Vpp and again the speaker cone will move +-5mm (10mmpp).

I imagine that this speaker will continue to move 10mmpp as the frequency is increased to audio frequencies and continue to do so through some of the audio range.
A big exception is the range around resonance when the movement will increase substantially.

At high frequency the movement will decrease from the standard 10mmpp.
What is the hf displacement reduction frequency? I suspect it is related to energy absorbed by accelerating the cone and the inductance of the VC. Any predictions?

[Ron E]

Originally posted by AndrewT
Hi,
I am kind of new to this speaker lark, but that does not sound right.

Svante and sreten know what they are speaking of.

At DC, the only thing that matters is stiffness, at ~3x resonance and up all that matters is mass, neglecting inductance. At resonance, the contributions of stiffness and mass are equal, and damping dominates.

1/f^2 is correct - every time you halve frequency, excursion increases by 4 for the same SPL. The reason sealed boxes don't double with each halving below resonance is because the 12dB/oct rolloff reduces excursion requirements.

A speaker that had constant excursion with frequency would have a 12dB/oct rising response....

The excursion formula depends on whether the speaker is enclosed or not and what sort of enclosure there is. In free air, it acts the same as if it were in a very large box - like 10000 cubic feet or so so you can enter that into a program that calculates escursion.

[Nichol1997]

I have WinISD beta but it does not model the excursion of the driver. I just downloaded the pro version but it is asking for parameters of the driver that I don't have.

Is there a formula for excursion vs frequency for a ported enclosure? It is tuned to 14 Hertz. I know that around the tuning frequency the cone of the driver won't be moving much and most of the SPL will be coming from the port. So there probably is a complex relationship between the two, but for higher frequencies I would imagine there is a formula.

[AndrewT]

1/f^2 is correct - every time you halve frequency, excursion increases by 4 for the same SPL.

That sounds right.

Lets say if a subwoofer was held in free-air and the amplitude was constant and the frequency was swept from 100 Hertz down

he is asking about sweeping a constant amplitude and in free air. That is NOT constant SPL.
I think the three of you are talking about a constant SPL formula when the enquirer is asking about sweeping a constant signal voltage.

[sreten]

Originally posted by AndrewT

I think the three of you are talking about a constant SPL formula when
the enquirer is asking about sweeping a constant signal voltage.

Hi,

My assumption is he meant constant SPL with a theorectical
driver that has a flat response and constant voltage drive.

We have all stated the above does reflect a practical reality.

/sreten.

[sreten]

Originally posted by Nichol1997
I have WinISD beta but it does not model the excursion of the driver. I just downloaded the pro version but it is asking for parameters of the driver that I don't have.

Hi,

tick the "auto calculate unknowns" box at the bottom of the editor.

/sreten.

[Svante]

Maybe this graph helps of a butterworth aligned closed box helps?



The black curve is the response of the system, the red is the cone amplitude. Above the cutoff frequency (50 Hz) the cone amplitude is tilted by -12 dB/oct but the response is flat. Below the same frequency the response tilts by 12 dB/oct and the cone amplitude is flat.

It is at the low frequencies that the intiution works best.