Mechanical Aspects

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

Hi Mr. Pierce
You seem quite knowledgeable on the aspects discussed here. Do you have
an expression for the mechanical power factor of a driver? TIA
 
G

Guest

Guest
Archived from groups: rec.audio.tech (More info?)

Northstar wrote:
> Hi Mr. Pierce
> You seem quite knowledgeable on the aspects discussed here. Do you
have
> an expression for the mechanical power factor of a driver? TIA

Please explain what you mean by "mechanical power factor."

"Power factor," in normal EE terms, means essentially cosine of
the phase angle of the load.

What do you mean by the term in this context?
 
G

Guest

Guest
Archived from groups: rec.audio.tech (More info?)

<dpierce@cartchunk.org> wrote

>> Hi Mr. Pierce
>> You seem quite knowledgeable on the aspects discussed here. Do
>> you
> have
>> an expression for the mechanical power factor of a driver? TIA
>
> Please explain what you mean by "mechanical power factor."
>
> "Power factor," in normal EE terms, means essentially cosine of
> the phase angle of the load.
>
> What do you mean by the term in this context?
>

Must be the same thing? A perfect damper has a power factor of 1; a
perfect spring or flywheel has a power factor of 0, assuming cos
phi.

A simulation model of a speaker may have a stage each for the
electrical, mechanical and acoustic loads. That would be useful for
illustrating the general idea, but I don't know how it maps to the
actual mechanics of the speaker. I guess it would just be the
suspension springs and dampers.

cheers, Ian
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <1112142821.990423.125280@o13g2000cwo.googlegroups.com>,
dpierce@cartchunk.org says...
>
>
>Northstar wrote:
>> Hi Mr. Pierce
>> You seem quite knowledgeable on the aspects discussed here.
>> Do you have an expression for the mechanical power factor
>> of a driver? TIA
>
>Please explain what you mean by "mechanical power factor."
>
>"Power factor," in normal EE terms, means essentially cosine of
>the phase angle of the load.
>
>What do you mean by the term in this context?


The ratio of real mechanical power to apparent mechanical power,
i.e. the ratio of usable mechanical power that can do work, to
applied or apparent mechanical power.
 

Andy

Distinguished
Mar 31, 2004
147
0
18,630
0
Archived from groups: rec.audio.tech (More info?)

You can derive an expression from an equivalent electrical circuit as
Ian suggests. You will need to know quite a lot about the mechanics of
your drive unit. You will also have to approximate somewhat to get a
tractable expression for the acoustic impedance.

A worked example is given in "Fundamentals of Acoustics" by Kinsler and
Frey (it has now picked up Coppens and Sanders as authors and been
revised but the relevant pages appear to be still there according to
the contents page on Amazon).
 

Andy

Distinguished
Mar 31, 2004
147
0
18,630
0
Archived from groups: rec.audio.tech (More info?)

The chapter Transduction looks from the titles as if it contains the
main information. However, things have clearly been moved around, a
number of new chapters added and the terminology changed somewhat. My
copy will be whatever was current in the late 70s but I do not have it
at home. The worked example consisted of data for a drive unit for
which an equivalent electrical circuit was derived. The equivalent
acoustical elements were included using approximate expressions from an
earlier chapter deriving the acoustical impedance of a piston. The
efficiency was considered for constant voltage, current and power
input. This is from memory but I have a reasonable level of confidence
I am not muddling things up.

I am not sure what ratio you are after because I am not sure what you
mean by apparent mechanical power. What do you want the ratio to
express? The useful work done is against the acoustical resistance. The
supply is the electrical work. But you are not after the overall
efficiency but something involving just the mechanical aspects?
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <1112298509.985729.316840@l41g2000cwc.googlegroups.com>,
andy19191@fsmail.net says...
>
>You can derive an expression from an equivalent electrical circuit as
>Ian suggests. You will need to know quite a lot about the mechanics of
>your drive unit. You will also have to approximate somewhat to get a
>tractable expression for the acoustic impedance.
>
>A worked example is given in "Fundamentals of Acoustics" by Kinsler and
>Frey (it has now picked up Coppens and Sanders as authors and been
>revised but the relevant pages appear to be still there according to
>the contents page on Amazon).


Thanks. I have the third edition of "Fundamentals of Acoustics" with
authors Kinsler, Frey, Coppens, amd Sanders, but find no worked
example in it, so Amazon may be using contents of an earlier version.
Which edition was that or can you email or post the equation?
biwatace@hotmail.com (remove the high card).
If it is just Rm/Zm as specified in the third edition, I have that,
but Kinsler never specifies if Rm includes (Bl)^2/Re that I can find.
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

>earlier chapter deriving the acoustical impedance of a piston. The
>efficiency was considered for constant voltage, current and power
>input. This is from memory but I have a reasonable level of confidence
>I am not muddling things up.
>
>I am not sure what ratio you are after because I am not sure what you
>mean by apparent mechanical power. What do you want the ratio to
>express? The useful work done is against the acoustical resistance. The
>supply is the electrical work. But you are not after the overall
>efficiency but something involving just the mechanical aspects?


In the electrical domain, heating power of I^2 Re can be useful,
here it is not. *Useful* acoustic power/electrical input power is of
course efficiency with a driver. *Usable* includes heat, but the
I^2 Re part is not *useful* and is wasted in this case. But to answer
your question, mechanical or electrical or a combination is OK, I
was just interested in mechanical power expressions.
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <iN_2e.7839$Nr5.7756@fe2.news.blueyonder.co.uk>,
IanIveson.home@blueyonder.co.uk says...
>
>Don't know where this came from
>
>http://www.ivesonaudio.pwp.blueyonder.co.uk/driver1.gif
>
>Just to give an idea.
>
>Should the suspension be critically damped? Can PF be derived from
>Q?
>
>cheers, Ian


Maybe so, but I believe the relationship would then be frequency
dependent, i.e apply only at mechanical resonance.

The website you noted is an equivilant circuit (very complete, note),
but does not get into the mechanical power aspect. Thanks.
 

Andy

Distinguished
Mar 31, 2004
147
0
18,630
0
Archived from groups: rec.audio.tech (More info?)

The rate of work is frequency dependent for the acoustic, mechanical
and electrical components. It does not apply just at resonance but at
all frequencies. What Ian has posted is a model of a full loudspeaker
from which you can extract anything you want. If that is frequency
independent information then you will have to integrate over the
frequency range of interest.

I am in the office but someone has run off with my copy of Kinsler and
Frey. However, the same stuff is in "Acoustics" by Beranek and, I
suspect, most of the engineering orientated Acoustics texts from the
40s, 50s and 60s.
 
G

Guest

Guest
Archived from groups: rec.audio.tech (More info?)

Northstar wrote:
> In article <1112343952.736126.293820@o13g2000cwo.googlegroups.com>,
> andy19191@fsmail.net says...
>
> >What Ian has posted is a model of a full loudspeaker
> >from which you can extract anything you want.
>
> Then how do you derive the mechanical power factor from the model?

The model posted is pretty typical of the elctrical model as
seen by the amplifier. You can break the model shown into three
parts:

1. Voice coil electrical
2. Mechanical system
3. Acoustical radiation

The sketch as shown does that neatly for you. WHat's the problem,
then, with analyzing each portion as you need to?

> Beranek gives no expression for mechanical power factor that
> I have found, nor does Olson, Morse et al.

There is no exact closed form expression, per se, for mechanical
power factor: you can derive it empirically from the from the
phase angle of the impedance of the mechanical leg of the circuit.
So, why not take the model you see, run it through something
like SPICE. COnsider:

# B200 impedance model

.SUBCKT B200 1 10
#
# Voice coil model
#
Re 1 2 9
Le1 2 3 0.04MH
Re2 3 4 19
Le2 4 5 1MH
#
# Mechanical model
#
Cmes 5 10 1400UF
Mces 5 10 5MH
Res 5 10 15
#
# Acoustic model
#
RRa 5 7 13.3
CRa 7 10 14UF
.ENDS

So, run a sweep, measure the phase across nodes 5 and 10
re the input phase, and take the cosine of that. That IS
the mechanical power factor vs frequency of the (in this
this case) driver.

In a simple model like this, yes there is an expression
for it, but this represent a VERY simple model of a driver
with totally linear mechanical properties and no interaction
with a cabinet and such. But, despite that, it'll get you
well in the ballpark.

Why do you want to know this?
 
G

Guest

Guest
Archived from groups: rec.audio.tech (More info?)

No. because, as I said there is no one "expression." It all depends
upon the specific details of the model. Instead, I provided you with
almost the entire means of getting the frequency-dependent mechanical
power factor.

If you want to take a given model and dervice an expression from it, go
for
it. It's not a trivial excercise given the complexity of the model.
Essentially,
you have to dervice the complex transfer function for the mechanical
portion of the model, and solve it for phase vs freuqency.

If you're looking for a single number that says "the mechanical power
factor is x", forget it. There is no such single number. It is, in the
simplest
of models, a co,pex frequency-dependent function. You can get a single
number.

Now, when I asked why you wanted to know this, being told that you're
writing
an article doesn't tell me much. The question is, why do you think this
information is important, in the sense that there might be a better way
of
trying to understand what you're doing.

In other words, what use is having a numver, frequency dependent as it
is
of mechanical power factor?
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <1112343952.736126.293820@o13g2000cwo.googlegroups.com>,
andy19191@fsmail.net says...


>What Ian has posted is a model of a full loudspeaker
>from which you can extract anything you want.

Then how do you derive the mechanical power factor from the model?

>I am in the office but someone has run off with my copy of Kinsler and
>Frey. However, the same stuff is in "Acoustics" by Beranek and, I
>suspect, most of the engineering orientated Acoustics texts from the
>40s, 50s and 60s.

Beranek gives no expression for mechanical power factor that
I have found, nor does Olson, Morse et al.
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <1112376682.030996.191980@f14g2000cwb.googlegroups.com>,
dpierce@cartchunk.org says...
>
>
>Northstar wrote:
>> In article <1112343952.736126.293820@o13g2000cwo.googlegroups.com>,
>> andy19191@fsmail.net says...
>>
>> >What Ian has posted is a model of a full loudspeaker
>> >from which you can extract anything you want.
>>
>> Then how do you derive the mechanical power factor from the model?
>
>The model posted is pretty typical of the elctrical model as
>seen by the amplifier. You can break the model shown into three
>parts:
>
> 1. Voice coil electrical
> 2. Mechanical system
> 3. Acoustical radiation
>
>The sketch as shown does that neatly for you. WHat's the problem,
>then, with analyzing each portion as you need to?
>
>> Beranek gives no expression for mechanical power factor that
>> I have found, nor does Olson, Morse et al.
>
>There is no exact closed form expression, per se, for mechanical
>power factor: you can derive it empirically from the from the
>phase angle of the impedance of the mechanical leg of the circuit.
>So, why not take the model you see, run it through something
>like SPICE. COnsider:
>
> # B200 impedance model
>
> .SUBCKT B200 1 10
> #
> # Voice coil model
> #
> Re 1 2 9
> Le1 2 3 0.04MH
> Re2 3 4 19
> Le2 4 5 1MH
> #
> # Mechanical model
> #
> Cmes 5 10 1400UF
> Mces 5 10 5MH
> Res 5 10 15
> #
> # Acoustic model
> #
> RRa 5 7 13.3
> CRa 7 10 14UF
> .ENDS
>
>So, run a sweep, measure the phase across nodes 5 and 10
>re the input phase, and take the cosine of that. That IS
>the mechanical power factor vs frequency of the (in this
>this case) driver.
>
>In a simple model like this, yes there is an expression
>for it, but this represent a VERY simple model of a driver
>with totally linear mechanical properties and no interaction
>with a cabinet and such. But, despite that, it'll get you
>well in the ballpark.

Thanks.


>Why do you want to know this?


For an article I am writing. Care to post the expression? TIA
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <1112394887.468090.248090@l41g2000cwc.googlegroups.com>,
dpierce@cartchunk.org says...
>
>No. because, as I said there is no one "expression."


Sorry, but you are wrong. Mechanical power factor is simply the
ratio of mechanical resistance to mechanical impedance, i.e.
PFmec=Rmec/Zmec. This is made clear in chapter one of
"Fundamentals of Acoustics" by Kinsler et al.


>It all depends upon the specific details of the model.


*All* worthwhile models should have a derivable mechanical
resistance and mechanical impedance.


>Instead, I provided you with almost the entire means of getting
>the frequency-dependent mechanical power factor.


Thank you, but details were not clear, and you clipped them out this
time around. My interest was in expressions for PFmec using parameters
other than resistance and impedance.


>If you want to take a given model and dervice an expression from it,
>go for it. It's not a trivial excercise given the complexity of the
>model. Essentially, you have to dervice the complex transfer function
>for the mechanical portion of the model, and solve it for phase vs
>freuqency.
>
>If you're looking for a single number that says "the mechanical power
>factor is x", forget it. There is no such single number. It is, in the
>simplest of models, a co,pex frequency-dependent function. You can get
>a single number.
>
>Now, when I asked why you wanted to know this, being told that
>you're writing an article doesn't tell me much. The question is, why
>do you think this information is important, in the sense that there
?might be a better way of trying to understand what you're doing.
>
>In other words, what use is having a numver, frequency dependent as it
>is of mechanical power factor?


Having PFmec makes the derivation of many other parameters simple
and intuitive. Sometimes an angel resides in the details as well
as the devil. :)
 
G

Guest

Guest
Archived from groups: rec.audio.tech (More info?)

"Northstar" <xxx@yyy.zzz> wrote

>>No. because, as I said there is no one "expression."

> Sorry, but you are wrong. Mechanical power factor is simply the
> ratio of mechanical resistance to mechanical impedance, i.e.
> PFmec=Rmec/Zmec. This is made clear in chapter one of
> "Fundamentals of Acoustics" by Kinsler et al.

Exactly. That's what we have been showing you.

PF = cos phi = R/Z.

The point of the electrical model is that its transfer function is
the same as that of the mechanical system. You just need to convert
units. Is that the part you are having difficulty with?

You will end up with an expression containing stiffness, viscosity,
mass and frequency. Is that what you want?

cheers, Ian
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <28H3e.15881$C12.4075@fe1.news.blueyonder.co.uk>,
IanIveson.home@blueyonder.co.uk says...
>
>"Northstar" <xxx@yyy.zzz> wrote
>
>>>No. because, as I said there is no one "expression."
>
>> Sorry, but you are wrong. Mechanical power factor is simply the
>> ratio of mechanical resistance to mechanical impedance, i.e.
>> PFmec=Rmec/Zmec. This is made clear in chapter one of
>> "Fundamentals of Acoustics" by Kinsler et al.
>
>Exactly. That's what we have been showing you.
>
>PF = cos phi = R/Z.


Yes that's basic, but how did you derive R and Z from the KEF analogous
circuit you referenced? Also how did you derive phi?


>The point of the electrical model is that its transfer function is
>the same as that of the mechanical system. You just need to convert
>units. Is that the part you are having difficulty with?


No, Cmes=M/(Bl)^2, Lces=(Bl)^2/s, Res=(Bl)^2/Rm


>You will end up with an expression containing stiffness, viscosity,
>mass and frequency. Is that what you want?
>
>cheers, Ian


Then what is that expression as you see it? TIA

What I want is expressions for mechanical power factor using
terms other than mechanical resistance and mechanical impedance.
 

Andy

Distinguished
Mar 31, 2004
147
0
18,630
0
Archived from groups: rec.audio.tech (More info?)

> Then what is that expression as you see it? TIA

Unless we know what you want this ratio to express this question is not
answerable. The reason being that unless you are clear what physics is
in and what physics is out you cannot extract what you want from the
posted equivalent circuit.

I am not having a go but my experience dealing with engineers in
industry is that they often ascribe properties to these "performance
factors" they do not really possess.

> What I want is expressions for mechanical power factor using
> terms other than mechanical resistance and mechanical impedance.

Since your power factor is a ratio (nondimensional) the units will not
matter.
 

northstar

Distinguished
Oct 21, 2004
15
0
18,560
0
Archived from groups: rec.audio.tech (More info?)

In article <1112687524.999637.201010@g14g2000cwa.googlegroups.com>,
andy19191@fsmail.net says...
>
>> Then what is that expression as you see it? TIA
>
>Unless we know what you want this ratio to express this question is not
>answerable. The reason being that unless you are clear what physics is
>in and what physics is out you cannot extract what you want from the
>posted equivalent circuit.
>
>I am not having a go but my experience dealing with engineers in
>industry is that they often ascribe properties to these "performance
>factors" they do not really possess.


Assigning a given voltage to the circuit input produces a current and
therefore a mechanical force BlI. This force then creates motion with
a velocity of force/mechanical resistance. The extent that force is in
phase with velocity then is the mechanical power factor.

Where is DP when we need him here? :)


>> What I want is expressions for mechanical power factor using
>> terms other than mechanical resistance and mechanical impedance.
>
>Since your power factor is a ratio (nondimensional) the units will not
>matter.
 

ASK THE COMMUNITY