Bill W. said:
I simplifying your math into a magnitude to be apparent
in my later equation. I have ask you to include
magnitudes for better communication, but you continue
to omit them. Why is that?
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Excuse me? I did post the magnitudes that you asked for. In addition, every
time that I used the phasor form, I expressed it in both rectangular and
polar forms. The polar form, as I explained, is (magnitude) @ angle so the
magnitude is there loud and clear.
Anyhow, I repeat what I sent (and it was sent) below.
************
Actually the magnitudes of the terms are given in my original calculation. I
showed the polar form which is (magnitude) @ (angle) as well as rectangular
form (real) +j(reactive)
For data I used only the values of E=1.41 volts, Re=5.78 ohms, Bl=7.17
Newtons/Ampere or volt-sec/meter Rms=2.23 N-s/m, M=0.0253Kg and K=3425 N/m
as I had indicated.
The following were calculated
..
V=0.0492 @ -71.76 degrees magnitude 0.0492 m/s
Zm =33.82 @ 86.22degrees magnitude 33.82 N-s/m (NB I did have the
angle wrong before)
Fm =1.664 @ 14.46 degrees magnitude 1.664 N
F=(Bl)E/Re =1.749 N equivalent source force with all electrical side
referred to mech side.
(Bl)^2/Re=8.89 N-s/m Electrical resistance referred to mechanical side.
In addition, as you requested:
F-((Bl)^2/Re)V =Fm =1.749-(8.89*0.492 @-71.76) =1.749-0.1369 +j0.415
=1.612+j 0.415
=1.665 @ 14.5 degrees magnitude 1.665N
This agrees with Fm calculated using ZmV (as it should )
************
Going a bit further, I =Fm/Bl =0.232 @ 14.46 degrees magnitude 0.232A
Eb =(Bl)V =0.353 @ -71.76 degrees Magnitude 0.353 volts
check: E-RI=1.41 -(5.78*0.232) @ 14.46 degrees =0.353 @71.6 degrees OK
The results are consistent and do satisfy the original basic equations
Note I is not 1.75/7.17 =0.244 magnitude (which is E/Re independent of
frequency and mechanical load)
-
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Sorry. I said I came here to learn, didn't I?
in
Ludwig's site which gives the equivalent circuit.
I am not going by Luswigs model. Never was.
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I realise that - but he has put it in a nutshell (and has based his work on
Small etc)
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Rme = (Bl)^2/Re is included as a resistive (therefore
lossy) term in models by Small, Beranek, Morse, Colloms,
Olson, Keele, Villchur, Kinsler, Kloss, and Lahnakoski.
It is used as an *equivilant* mechanical resistance,
and their models are all essentially alike. Why do
you use Ludwigs model?
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I don't have the other's and Ludwig's model, based on Small's work presents
things in a way that is, to me, fairly intuitive. From what you have quoted,
it appears that Small and others actually have used a similar model to
develop their final simplified models.
I have no problem with it being an "equivalent mechanical resistance" and
have said as much.
"This becomes F=Fr +Fmec where Fr is the (Bl)^2(V/Re isbelow)
However, it is a reflection of the electrical resistance and is not part of
the actual mechanical impedance which is due to only the mechanical
elements. It is useful in analysis.>
----------------------
at.
Do all the math you like, but as I noted, it is
applied mechanical force
--------------
Bull. It is a electrical current source equivalent expressed in mechanical
terms.
Does representation of the mechanical parameters in terms of equivalent
electrical parameters make them actual electrical elements and with losses
that are no longer mechanical but electrical? I could bring Rms over to the
electrical side (it will be 23 ohms ) and then claim that there are no
mechanical losses -just electrical losses. Do you see the flaw in this? Just
because the mechanical resistance can be expressed in terms of an electrical
resistance, does it follow that the losses are then actually electrical
losses and not mechanical losses at all?. That is illogical . Yet your
argument shows the same illogic.
Also note that from the force you have given, I =1.75/7.17 =0.244 and this
is the same as E/Re =1.41/5.78 =0.244 A Is this the actual current ? Also
this current and the corresponding force are independent of the mechanical
impedance and the frequency. Does that make sense to you? Do you expect
the force and the current associated to be constant? Is the current at
resonance the same as the current at 227Hz? If not then there must be
something wrong with the assumption that the actual mechanical force is
1.75N.
Alternatively the "force" of 1.75 N could possibly be other than the actual
mechanical force applied to the mechanical system. That is my position based
on straight- forward circuit analysis (incidently no phasors needed for
this ) .
Back to basics
We have E-RI =Eb for the electrical system where Eb is the back emf =(BL)V
We also have Fmec =ZmecV where Zmec depends on the various masses, spring
compliances and damping factors of the mechanical system. and V is the
mechanical velocity This expression deals only with the actual mechanical
dynamics of a force acting on a mass, spring,damper system and does not deal
with any electrical factors per se. .
(In this case Zmec =Rms +j(wM-K/w) = sqrt(Rms^2 +(wM-K/w)^2) @ angle arctan
(wM-K/w)/Rms )
These are the basic equations which you seem to have accepted If you haven't
then it is back to square one.
Rewrite the electrical equation
(E-Eb)/Re =I
In addition Fmec =(Bl)I
so Fmec =(Bl)I =(Bl)E/Re -(Bl)Eb/Re =ZmecV
but (Bl)Eb/Re = (BL)^2V/Re
so Fmec ={(Bl)E/Re -(Bl)^2V/Re} = ZmecV ***
or (Bl)E/Re =[(Bl)^2/Re +Zmec]V
(Bl)E/Re is not the actual mechanical force (Fmec ) but it as well as the
(Bl)^2/Re term make up a non-ideal current source referred to the mechanical
side. This approach is convenient for determining V and from V , Fmec can be
found. From V and Fmec, the actual mechanical power can be found as well as
the electrical current and the back emf.
------------------
You still dn't acknowledge that 0.0054 is not the
total net mechanical power???
------------
I certainly don't and I have given my reasoning for that. You have not given
reasoning for your contention. You have made quotes but the one you have
given re (BL)^2/Re supports my contention- not yours. I have stated that it
is the electrical resistance term referred to the mechanical side - as such,
even though it is an "equivalent mechanical resistance" it is not an actual
mechanical resistance. There is a very important difference.
-----------------
This is not correct as I have shown. Again, do
all the math you like, but here's the bottom line.
Net mechanical power is just velocity squared times
system *resistance*, analogous to electrical
where P = I^2 R.
---------------------
First of all, you haven't shown anything. You have made assertions without
providing the reasoning for your stand. Sorry but that doesn't cut it. When
you quote references, your quotes support my position as well as, or better,
than they support your position.
Since the basis of any model of an electromechanical transducer is, of
necessity, for working purposes, mathematical it is hard to avoid using
mathematical methods- hand waving doesn't work. (Small, et al didn't avoid
mathematical methods in their developments). I have shown a different
approach to the basic equations ( above) . Go through it step by step, not
worrying about phasors. It is not difficult math (you have shown that you
can handle more difficult stuff) but it does show where the terms in
contention comes from. Then come back with questions about it.
Noting that the term (Bl)^2/Re depends on an electrical resistance, how can
it involve actual mechanical power? This term along with (Bl)E/Re make up
the equivalent current source "as seen from the mechanical side". Since
(Bl)^2/Re is simply a consequence of the model and the electrical resitance
and is not an actual mechanical element , the power into it , while
expressed in mechanical terms is not a an actual mechanical loss but a loss
in the electrical side equivalent to Eb^2/Re. Unfortunately this does not
translate into I^Re for reasons which I won't get into.
Pmec = v^2 * Rmec
= v^2 * (Rme + Rms)
= 0.0493^2 * (8.89 + 2.23) = 0.027
Perhaps this will help:
Generally power is noted as force times velocity,
if force and velocity are in phase, otherwise
Pmec = F v PF = 1.75 * 0.0493 * 0.313 = 0.027
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no kidding. I have nothing against the form of that equation but I do have a
problem with the values used. The power that you have given is the
mechanical power plus an internal loss in the current or force source.
----------
Sorry, but there is no point in my addressing your
following comments below until this is settled, as
to who is correct on the magnitude of net mechanical
power.
Others are welcome to comment on this.
Bill W.
--------------
Fair enough -but at least read what I have said and the basis for the
inclusion of the (Bl)^2/Re term as well as the difference between the
(Bl)E/Re "force" and the actual mechanical force.
Note that none of the "math" that I have used is anything special - It is
simply application of circuit analysis to an electromechanical system -
quite commonly done. It is also obviously what Small etc have also done. I
referred to Ludwig because it was the only source that I could find on the
net that actually had some meat and went back to get show the basis of
Small's work. I would love to see Small's original paper. The others seem to
involve unspecified programs that would provide a model when the data was
fed in - essentially someone has taken the math and used it to write a
program. Other sites give the way to make certain measurements and provide a
program to caculate the parameters.
--
Don Kelly
[email protected]
remove the urine to answer