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T

tmb

Jan 1, 1970
0
What other Forums or Newsgroups are on the net for engineers who specalize
in...

- Electronic Circuit Design
- System Design
- Antenna Design

Thanks for any info.
 
D

Don Kelly

Jan 1, 1970
0
----- Original Message -----
From: "Bill W." <[email protected]>
Newsgroups: alt.engineering.electrical
Sent: Thursday, December 11, 2003 11:15 AM
Subject: Re: Motor torque and back emf

Before I reply to your last post, would you please
give the magnitudes of your terms as used above, i.e:
V
Zm
Fm
F
I=E/Re
e
F=BlE/Re
(Bl)^2/Re
F-((Bl^2)/Re)V (using phasor as you note)

Not nit-picking, just trying to correlate terms, and
thanks again for all.

TIA

Bill W.
---------
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 @ -71.76 degrees magnitude 33.82 N-s/m
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 )

Note none of the calculated values are actually as accurate as shown.
3 significant digits is all the data will allow so on that basis Fm=1.67,
F=1.75, Zm =33.8 in magnitude and further digits are really meaningless. In
fact even 3 significant digits is probably excessive as I doubt whether some
of the measured data is actually that good.

Don Kelly
[email protected]
remove the urine to answer
 
D

Don Kelly

Jan 1, 1970
0
Bill W. said:
------ shipping ---------


Bill W wrote:







How did you arrive at 14.46 degrees here? TIA
---------------------------
I calculated the velocity from BlE/Re =Zme V (Zme is the total impedance
referred to the mech side.
Then I looked at Fm=VZm where Zm is the mechanical impedance. and found the
corresponding I
Alternatively, I get the same result from E-RI =BlV =ZmeI where Zme is the
mechanical impedance transferred to the electrical side. (or E=ZeI where
Ze =Re +Zme)
These calculations were given.The latter is
1.41 =(Re+(Bl)^2/[Rms+j(wm-K/w)]I =(5.88-j1.517)I giving I =0.232 @14.46
I showed the calculations and results. Shall I repeat them?
--------------------








Don Kelly followed up re the above:






Taking your equation VbZm =Fm = 1.664 @14.46 N ,
ammended by you as "Vb should have been V", gives

V Zm =Fm = 1.664 @14.46 N,

but

V Xmec = 1.664, not V Zmec

V Xmec = V wm-k/w = 0.0493*33.75 = 1.664
----------
I stand by what I said. Zm=2.32+j33.75 =33.82 @86.22 degrees
V=0.0492 @-71.76 degrees
The product F=1.664 @14.46 degrees
The difference between 33.75 and 33.82 is 0.2% in magnitude.
You also are using V=0.493 where I calculated a magnitude of V of
0.0492 -this is part of the difference. Do you really know V to an accuracy
of 5 parts in 10,000 as implied?, or even 1 part in 1000? How about 3% which
puts F somewhere between 1.61 and 1.71?
You are approximating Zm by using Xmec and only calculating the magnitude
of the force. Using your approach with |V|=0.0492 (my value calculated from
the base data) I get 1.660 N magnitude. Since the phase angle is large, this
approximation is good for the magnitude of the force.
As I said before, the data doesn't support 4 figures so 1.66 is as good as
you get for magnitude.
-----------
So here is where I have a problem. Your Fm equation
which you define as "Fm is the actual mechanical force"
is just the Lorentz force of

BLI = 7.17*.232 = 1.66344 (using your I of 0.232)
----------
Yes-it is the Lorentz force -- except that I have taken into account the
phase.
F=1.66 N. Saying 1.66344 is garbage.- you can't squeeze 6 figure accuracy
out of 3 figure data even if your calculator says so.
---------------
Matching your Fm of 1.664 within 0.03%. Not likely
to be coincidence, but regardless I ran the figures
on all 12 drivers equating

V Xmec = BLI

V Xmec matches BLI on the 12 drivers within an average
of 2.4%. Not bad, as many consider 3% matching to be
excellent due to measurement tolerances.
------------
See above (and at the end) - VXmec gives a good approximation for magnitude
at this and higher frequencies. What it does is use the approximation Zm
=jXmec =Xmec @ 90 degrees. This will lead to about a 4 degree error in the
phase of I
-----------------
Now your equation as equated to BLI is correct at
resonance where current I and voltage E are in phase,
but I do not think it should be applied at 227.4 Hz,
since I is not in phase with E at 227.4 Hz.
-----------------
The whole purpose of my approach is because there are phase shifts and the
approach takes these into account. At resonance it is not necessary simply
because things are in phase. At any other frequency there are phase shifts
and any approach used must account for these. Using the phasor methods as I
have done is the normal method of AC circuit analysis as applicable in this
case.
Using magnitudes only is something that gets beaten out of students in their
first AC circuits course.

My use of the equations is correct at any frequency for which the data is
valid. If the initial data is correct, the results will be correct. In
fact, what I have been trying to tell you is that, at 227.4Hz your approach
approximates the magnitudes (and gives a good aproximation at this and
higher frequencies) but doesn't account for phase effects.
Working with magnitudes only causes problems.
For example you had discrepancies in Eb as calculated from E-RI and from BlV
because you worked only with magnitudes. - using the phasor methods gives
consistent results as E-RI =Eb =BlV bloody well better be true whichever
way you calculate it -if not it is a red flag that something is very wrong.
------------------------
Regardless, your equation would give zero actual or net
force at resonance, and again we have no power Captain..

At resonance

V Xmec = V wm-k/w = 0.157*zero = no force
-----------------
YOUR equation implies the above- this is in contradiction to your own data.
VXmec is YOUR approximation, not mine, and it breaks down here. Please do
not assign your errors to me as I am quite capable of making my own.

At resonance:

Zm =Rms+j0 at resonance so F=0.157(2.23)=0.35 N (at angle 0) {*****doesn't
look like 0}
Note that this agrees with the following
Ze=Re+(Bl)^2/(Rms +j0) =5.78+23.05 =28..83 ohms (angle 0) (electrical
input impedance)
I=1.41/28.83=0.0489 A and corresponding F=7.17*0.0489 =0.351 N *****
Also Eb =1.41*23.05/28.83=1.127v (by voltage divider) so V=1.127*7.17=0.157
m/s ****
(or Eb=1.41 -5.78*(0.0489+j0)=1.127V at angle 0)****
---------------------------
So unless I've missed something here, V*Xmec is not
the expression for actual or net force, and the
magnitude of net force at 227.4 Hz is not 1.664.
---------------
V*Xmec is not the actual force. V*Zm as a phasor quantity is.
As for magnitude, I calculated a magnitude of 1.664 but recognise that 1.66
is as meaningful value as you will get.
Zm has a magnitude of 33.82 Ns/m Xmec =33.75 N-s/m which is within 0.2% of
Zm at this frequency. What you are effectively doing is saying Zm
=0+j33.75 =33.75 @90 degrees as opposed to the true value of 2.23 +j33.75
=33.82 @86.22 degrees. It is a good approximation but using this approach at
other frequencies leads to errors. In addition, with correct magnitudes and
no phase information, further calculated results are nonsense.
Using arithmetic based on magnitudes only can, and generally will, give
incorrect results for AC conditions. The authors of your references
recognise this but were trying to set up a simplified approach for a given
purpose, recognising where the approximations are valid and also where they
are not.
 
D

Don Kelly

Jan 1, 1970
0
Bill W. said:
-------------



Your reply is below, unclipped.

You defined Fm as "Fm is the actual mechanical force",
and go on to state:
"The mechanical power is 1.664(0.0492)cos (14.46--71.76)
=5.4mW That is the mechanical power factor is used.
or more easily Pm=Rm |V|^2 =5.4mW",

This gives mechanical power Pmec as equal to velocity
multiplied by the terms other than velocity
-----------------
So? Velocity has to be multiplied by something other than velocity to get
power.
Rm is N-s/m so Pm is given in (N-s/m)(m/s)^2 =Nm/s which is power The
units are OK. I checked them before.
--------------
Pmec = 0.0492 * 0.11 = 0.0054 watt or 5.4 mw

I'll get to the 0.11 and 0.0054 magnitudes below.

Now I have repeatedly said this represents only PART
of the mechanical power and/or force, in other words
this is just the power dissipated in overcoming the
suspension resistance, not the total mechanical power.
Instead of addressing this directly, you have now
chosen to be derisive. So be it.

Halliday et al, 6th ed, 7-47 gives mechanical power as

Pmec = F cos angle v
------------
No problem with Halliday but a problem with what you "repeatedly" (and
incorrectly) said.
In the problem as given, there is no other mechanical loss term. The Re term
is not associated with mechanical loss. I have said this before. . So far we
have only been considering a model of the driver (the model as indicated in
Ludwig's site which gives the equivalent circuit. The only lossy elements
are Re (electrical) and Rms (mechanical). Do you have data for other
mechanical elements such as those shown by Ludwig (Lceb, Cmeb, Rel which
are cabinet parameters)?
----------
where here:
F = E Bl/Re = 1.41*7.17/5.78 = 1.75
--------------
Here there is a problem. I explained before how this force was arrived at.
When you use F=BlE/Re you are taking the current source model (see Siskind
or below) which can be represented by E/Re in parallel with Re (as I
explained before). and moving this over to the mechanical equivalent to get
F=BlE/Re driving the mechanical impedance Zm in parallel with
(Bl)^2/Re The equation is then F=BlE/Re +{(Bl)^2/Re + Zm)V where Zm is the
mechanical impedance. This becomes F=Fr +Fmec where Fr is the (Bl)^2(V/Re is
due to the coil resistance as seen from the mechanical side. It is not an
actual mechanical element although it is represented by one. (going the
other direction, Rms converts to an electrical equivalent conductance
(Bl)^2/Rms )
----------------
cos angle = power factor = PF = 0.313
(as given by you), or
PF = Rmec/Zmec = 11.12/35.5 = 0.313
(as given by me)
v = 0.0493
------------------
Note that your pf=0.313 agrees with the phase angle (71.76 degrees) of the
total Zme = coil R moved to the mechanical equivalent ) +Zm where Zm =Rms
+j(Mw-K/w)
------------------
therefore

Pmec = F PF v = 1.75*0.313*).0493 = 0.027
-------------------
and here is the rub: The Pmec you have calculated includes the (Bl)^2/Re
term which is NOT part of the mechanical impedance. The F that you use is
not the actual mechanical force but the current source equivalent.
That is why I calculated the net mechanical force acting on the mechanical
impedance (1.664 N) from ZmV where Zm =(2.23 +j33.75) =33.82 @86.22

From this Pmec =2.23(0.0492)^2 =1.664*0.0492* (Pf) =5.4mW where
Pf=2.23/33.75=cos 86.22 =0.066
Halliday's expression is fine- (it is a standard expression) provided you
use the actual Zm.and actual mechanical force. It accounts for the phase.
He probably used the approach that I have used but expressed it in terms of
pf. I have no argument with Halliday regarding this.
---------------
Rearranging Hallidays equation of Pmec = F PF v

F*PF = Pmec/v = 0.027/0.0493 = 0.548

Note that power factor PF = cos angle in the F*PF
expression clearly ACCOUNTS FOR PHASE, such that
F*PF = 0.548 is the component of force in the
direction of velocity, i.e. the actual or net
force.

In other words

F net = Pmec/v = 0.027/0.0493 = 0.548

Now to relate net force to resistances

F net = (v Rms) + (v Rme)
= (0.0493*2.23) + (0.0493*8.89)
= 0.11 + 0.438 = 0.548
There is your 0.11 magnitude and as you can see, it
represents only PART of the net or actual force.
------

NO: Do you realise that you have just justified what I said?

Ftotal =Fc +Fm where Fm is the actual mechanical force and Fc a force which
is related to the electrical resistance Re.
Note that Eb/Re =0.0611A which corresponds to a "force" (Fc) of
0.0611(7.17) =0.438 N which is a result of the model used.
This leaves us with the force acting on the actual mechanical resistance
which is 0.11 N.
Also note that your force 1.75 N converts to a current of 0.244A (and the
equivalent electrical current source = E/Re is a current of magnitude
0.244A.) Is the actual current 0.244A? We agree that it isn't. 0.232A
corresponds to 1.663+ N which is what I claim for the actual mechanical
force.
At least what I have written is self consistent and differentiates between
the electrical resistance, as represented on the mechanical side, and the
actual mechanical resistance.

As to the current source model

Voltage source model Current source model
Re
|--------\/\/\-----------o
|------------|---------o
| + -->I + | +
| -->I
E Eb Is
Re Eb
|_________________o |_________|_______o

E-RI =Eb or E/Re =Eb/Re +I Is =E/Re =Eb/Re +I or E/Re
=Eb/Re +I
The two are equivalent *as seen at the Eb terminal*.

Is is not I. The difference is the current Eb/Re
The force BlE/Re is the current source (Is) transferred to the mechanical
side. It is not the actual force seen by the mechanical system as the Eb/Re
term becomes a ((Bl)^2/Re)V term and the "force" in this element is NOT
part of the actual mechanical force as this element is not one of the
mechanical elements.
----------------

where
Rms = (2 pi fc m/Qmc = 2*pi*58.6*0.0253/4.17 = 2.23
= suspension mechanical resistance.
Rme = (Bl)^2/Re = 7.17^2/5.78 = 8.89 = resistance
encountered in overcoming back emf, which Small
has called mechanical resistance of the motor,
or sometimes called effective resistance of the
motor.
------------------
Small is doing what I have done- simply expressed the electrical resistance
in equivalent mechanical terms. I repeat, Rme doesn't represent an actual
mechanical element. It represents the effect of coil resistance (NOT back
emf related) and it must be separated from the mechanical resistance when
calculating mechanical power. OK?

Also, I don't know whose terminology it is to say it is the "resistance
encountered in overcoming back emf". That is a crock. Someone is very sloppy
in their terminology.
Back emf is entirely due to the velocity of the motor. Nothing more, nothing
less.
It has nothing to do with Re. If the cone was clamped so that it could not
move, Re would still be there and a mechanical force would still exist but
Eb would be 0. If Re was 0 and the cone could move, then a force would
still exist and cause velocity and a back emf would exist.
Re has nothing to do with the back emf. It is a property of the
coil -nothing more. As such it enters into the force/velocity or
current/voltage relationships satisfying the fundamental equations of the
system. What I have written satisfies these equations - I haven't forgotten
that much circuit analysis or simple mechanical analysis.
----------
Note phasor addition is not required as the powers
are into resistance.
Now to net power:

Pmec = (v^2 Rms) + (v^2Rme)
= (0.0493^2*2.23) + (0.0493*8.89)
= 0.0054 + 0.0216 = 0.027

There is your 0.0054 power magnitude, and again as you
can see, it represents only PART of the net or actual
mechanical power.

Note this agrees EXACTLY with Halliday above..
----------
I agree with Halliday. You don't - as you have used a Z which includes the
electrical element Re
Note that the power in Re =|0.0216 =(Eb^2)/Re and is an electrical loss in
the current source model. It is NOT, repeat, NOT, an actual mechanical
power.
Also note phasor addition is not required here as well,
since this too is resistive.
-----------------------
No problem. . You are using |V|^2Rm correctly . Note however, that
Halliday, when using pf is in fact using phasor methods. In other
calculations, as I have shown, it is necessary to take into account phase
relationships - the use of phasors is simply a very convenient way to do so.
As you have shown by discrepancies, using magnitudes alone is not correct.
I'm sure that Halliday realises this. I haven't read Halliday nor am likely
to but I wonder how much explanation he gives for his formulae. Does he call
Rme an actual mechanical resistance or does he call it the mechanical
equivalent of the electrical resistance?
------------------
I shall send the above for peer review.
-------
Better wait. There are too many inconsistencies in your work or
interpretation of Halliday and others. If not, include my comments above.
------------
This constitutes my entire reply in this post, leaving
your derision to stand on its own.
-------------
I'm was not deriding you. I think that I am not getting through to you. I
have no reason to think that you are not good at what you do but I do think
that your understanding of the theory behind the analysis is weak and too
dependent on plugging in formulae without knowing the why of the equations
and the approximations involved.
Possibly Halliday and Small haven't made it clear as to the basis f their
development. That is why I referred you to the Ludwig site as he looked at
their work and showed the development of the model.

--
Don Kelly
[email protected]
remove the urine to answer



Bill W wrote:


Pmec = IEb = IE-I^2Re = 0.320 - 0.298 = 0.0222

Although it may require vector analysis to be broken
down in detail re power division, I believe it is
valid. See:
A.E. Fitzgerald et all (Electric Machinery) 6th ed.
p.389 on: Eb = E-IR (multiply by I for power)
Joe Kaiser (Electrical Power) 3rd ed. p.212:
IEb = IE-I^2R


Don Kelly wrote:

-------------------
No problem with using E=Ri +Eb for an AC situation
as long as the quantities are treated as phasors
(once called vectors). I did this:
E-IRe =1.41 - (5.78)0.232 @14.46 = 1.41 -1.34 @14.46
degrees =1.41-1.30 -j0.335 =0.11-j 0.335 =0.353 @-71.7
degrees (agrees) and the resultant Eb is the same as
that obtained from a mechanical analysis and with your
data.


How did you arrive at 14.46 degrees here? TIA
---------------------------
I calculated the velocity from BlE/Re =Zme V (Zme is the total impedance
referred to the mech side.
Then I looked at Fm=VZm where Zm is the mechanical impedance. and found the
corresponding I
Alternatively, I get the same result from E-RI =BlV =ZmeI where Zme is the
mechanical impedance transferred to the electrical side. (or E=ZeI where
Ze =Re +Zme)
These calculations were given.The latter is
1.41 =(Re+(Bl)^2/[Rms+j(wm-K/w)]I =(5.88-j1.517)I giving I =0.232 @14.46
I showed the calculations and results. Shall I repeat them?
--------------------


Don Kelly wrote:

The mechanical force is given by
VbZm =Fm = 1.664 @14.46 N


Bill W. wrote:

You lost me here. How are you defining the terms, and
please state the magnitudes. TIA
1. Vb
2. Zm
3. Fm


Don Kelly wrote:

------------
Vb - sorry -this should have been V - the mechanical
velocity. Zm is the mechanical impedance -mechanical
elements only as I have indicated above. Fm is the
actual mechanical force (F is the driving force as
a source treating it as the equivalent of a current
source I=E/Re in parallel with e -moviing to the
mechanical side the result is F=BlE/Re in parallel with
an equivalent mechanical resistance (Bl)^2/Re ) This allows
calculation of the velocity but to get the actual mechanical
force it is necessary to look at the velocity V times the
actual mechanical impedance Zm (or use F-((Bl^2)/Re)V
(using phsor arithmetic, not scalar). This corresponds
to a current of 0.232 @14.46 degrees (Amps) All the above
agree in magnitude with your values The mechanical power
is 1.664(0.0492)cos (14.46--71.76) =5.4mW
That is the mechanical power factor is used.
or more easily Pm=Rm |V|^2 =5.4mW


Don Kelly followed up re the above:

---------
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 @ -71.76 degrees magnitude 33.82 N-s/m



Taking your equation VbZm =Fm = 1.664 @14.46 N ,
ammended by you as "Vb should have been V", gives

V Zm =Fm = 1.664 @14.46 N,

but

V Xmec = 1.664, not V Zmec

V Xmec = V wm-k/w = 0.0493*33.75 = 1.664
----------
I stand by what I said. Zm=2.32+j33.75 =33.82 @86.22 degrees
V=0.0492 @-71.76 degrees
The product F=1.664 @14.46 degrees
The difference between 33.75 and 33.82 is 0.2% in magnitude.
You also are using V=0.493 where I calculated a magnitude of V of
0.0492 -this is part of the difference. Do you really know V to an accuracy
of 5 parts in 10,000 as implied?, or even 1 part in 1000? How about 3% which
puts F somewhere between 1.61 and 1.71?
You are approximating Zm by using Xmec and only calculating the magnitude
of the force. Using your approach with |V|=0.0492 (my value calculated from
the base data) I get 1.660 N magnitude. Since the phase angle is large, this
approximation is good for the magnitude of the force.
As I said before, the data doesn't support 4 figures so 1.66 is as good as
you get for magnitude.
-----------
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 )



So here is where I have a problem. Your Fm equation
which you define as "Fm is the actual mechanical force"
is just the Lorentz force of

BLI = 7.17*.232 = 1.66344 (using your I of 0.232)
----------
Yes-it is the Lorentz force -- except that I have taken into account the
phase.
F=1.66 N. Saying 1.66344 is garbage.- you can't squeeze 6 figure accuracy
out of 3 figure data even if your calculator says so.
---------------
Matching your Fm of 1.664 within 0.03%. Not likely
to be coincidence, but regardless I ran the figures
on all 12 drivers equating

V Xmec = BLI

V Xmec matches BLI on the 12 drivers within an average
of 2.4%. Not bad, as many consider 3% matching to be
excellent due to measurement tolerances.
------------
See above (and at the end) - VXmec gives a good approximation for magnitude
at this and higher frequencies. What it does is use the approximation Zm
=jXmec =Xmec @ 90 degrees. This will lead to about a 4 degree error in the
phase of I
-----------------
Now your equation as equated to BLI is correct at
resonance where current I and voltage E are in phase,
but I do not think it should be applied at 227.4 Hz,
since I is not in phase with E at 227.4 Hz.
-----------------
The whole purpose of my approach is because there are phase shifts and the
approach takes these into account. At resonance it is not necessary simply
because things are in phase. At any other frequency there are phase shifts
and any approach used must account for these. Using the phasor methods as I
have done is the normal method of AC circuit analysis as applicable in this
case.
Using magnitudes only is something that gets beaten out of students in their
first AC circuits course.

My use of the equations is correct at any frequency for which the data is
valid. If the initial data is correct, the results will be correct. In
fact, what I have been trying to tell you is that, at 227.4Hz your approach
approximates the magnitudes (and gives a good aproximation at this and
higher frequencies) but doesn't account for phase effects.
Working with magnitudes only causes problems.
For example you had discrepancies in Eb as calculated from E-RI and from BlV
because you worked only with magnitudes. - using the phasor methods gives
consistent results as E-RI =Eb =BlV bloody well better be true whichever
way you calculate it -if not it is a red flag that something is very wrong.
------------------------
Regardless, your equation would give zero actual or net
force at resonance, and again we have no power Captain..

At resonance

V Xmec = V wm-k/w = 0.157*zero = no force
-----------------
YOUR equation implies the above- this is in contradiction to your own data.
VXmec is YOUR approximation, not mine, and it breaks down here. Please do
not assign your errors to me as I am quite capable of making my own.

At resonance:

Zm =Rms+j0 at resonance so F=0.157(2.23)=0.35 N (at angle 0) {*****doesn't
look like 0}
Note that this agrees with the following
Ze=Re+(Bl)^2/(Rms +j0) =5.78+23.05 =28..83 ohms (angle 0) (electrical
input impedance)
I=1.41/28.83=0.0489 A and corresponding F=7.17*0.0489 =0.351 N *****
Also Eb =1.41*23.05/28.83=1.127v (by voltage divider) so V=1.127*7.17=0.157
m/s ****
(or Eb=1.41 -5.78*(0.0489+j0)=1.127V at angle 0)****
---------------------------
So unless I've missed something here, V*Xmec is not
the expression for actual or net force, and the
magnitude of net force at 227.4 Hz is not 1.664.
---------------
V*Xmec is not the actual force. V*Zm as a phasor quantity is.
As for magnitude, I calculated a magnitude of 1.664 but recognise that 1.66
is as meaningful value as you will get.
Zm has a magnitude of 33.82 Ns/m Xmec =33.75 N-s/m which is within 0.2% of
Zm at this frequency. What you are effectively doing is saying Zm
=0+j33.75 =33.75 @90 degrees as opposed to the true value of 2.23 +j33.75
=33.82 @86.22 degrees. It is a good approximation but using this approach at
other frequencies leads to errors. In addition, with correct magnitudes and
no phase information, further calculated results are nonsense.
Using arithmetic based on magnitudes only can, and generally will, give
incorrect results for AC conditions. The authors of your references
recognise this but were trying to set up a simplified approach for a given
purpose, recognising where the approximations are valid and also where they
are not.
 
D

Don Kelly

Jan 1, 1970
0
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?
---------------
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)
-
-------------------------------
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.
---------
I realise that - but he has put it in a nutshell (and has based his work on
Small etc)
----------
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?
----------
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
-----------
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]
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