To answer your question, re you completing the assessment of my analysis, before I respond:
All in steady state operation:
Mechanical: At 227.4 Hz, angle = arc cos Rmec/Zmec = 71.76 PFm = cos angle = 0.313 or Rmec/Zmec = 0.313
At 58.6 Hz, angle = arc cos Rmec/Zmec = 0 PFm = cos angle = 1.00 or Rmec/Zmec = 1.00
------
Electrical: At 227.4 Hz, angle = arc cos Re/Ze = 21.44 PFm = cos angle = 0.931 or Re/Ze = 0.931
At 58.6 Hz, angle = arc cos Re/Ze = 78.41 PFm = cos angle = 0.201 or Rmec/Zmec = 0.201
------ You have not yet responded to my question: Would you please state the *total* amplitude of power you see being transfered to the mechanical side at 227.4 Hz? Is it your magnitude of 0.0054 watt?
-------------- This includes the effect of Re as transferred to the mechanical side. If I use your Rmec as (Bl)^2/Re +Rms then the above is correct. As I have indicated before - this is an apparent Rmec driven by an apparent force.
Confusion due to my typo? Sorry. Also, I just now changed PFm to PFe above, to indicate electrical power factor.
Ok, the magnitude. So then, are saying the total power transferred to the mechanical side as net mechanical power is 0.0054 watt? If not, then what magnitude?
The total amount of *net* mechanical power in watts
Are you now using the angle of 86.22, otherwise what angle?
------------ I see where you get this "pf " but in fact, although it works for you, the power factor in the electrical sense, is 1 at resonance as Ze is purely resistive. The phase angle of Ze is 0. You have found a ratio Re/Ze (based on measurement of Ze?) =5.78/28.78 and called this a power factor. This really has no relationship to the electrical power factor as normally defined and can be misleading. It works for you. There is no phase angle involved in this case. It is simply a ratThis is not necessary as you can use Ze =Re +Zm as I indicate. It appears that you have defined power factor as the ratio Re/Ze which is not necessarily correct in the normal electrical use of PF where PF is based on the ratio of real to apparent power (watts/VA). This is the cosine of the angle between voltage and current and for an impedance Z=R +jX =root(R^2 +X^2) at angle arctan X/R then it is valid to use R/Z as you have done. However in that case the ratio is that of the real part of Z to the magnitude of Z. At resonance, Z is real =Re +23.0 =28.78 . current and voltage are in phase. and the real part of Z is Z, not Re.
Again- how are you determining an angle? You say angle =arc cos Re/Ze =78. 41 and pf =0.210=Re/Ze but at resonance you have simply the ratio of two resistors - what is important is the Re/Ze ratio -no need to associate a ficticious angle with this.
The nice thing is that it works at resonance but a bit of math shows that the result is the same as saying Ze =Re +(Bl)^2/Zm where Zm =Rms at resonance. At other frequencies the Zm will have both reactive and resistive components- the expreesion is good at all frequencies.
---------- Yes-as I said this is the magnitude of the average *real* power in watts It is not the "apparent power" in VA, the "reactive power" in VAR's or the peak power (instantaneous watts) which are the only other "powers" that I can see as being of interest.
I will note the following now, as an intermediate response re your last post:
Yes, I can see that view. The equation I gave above expresses one view that real power is the power associated with resistance. It's often stated that the power factor of a resistive heater (without motor or reactive element) is 1, and that 100% of the applied power is real. I would use the word useful, and I believe the word useful becomes *key* here, perhaps even in all cases. In this case, an electrical power factor of 1 at resonance gives useful power of (note the following is all referred to resonance)
Pin * PF = IE * 1 = 0.049*1.41*1 = 0.0691
Showing the IE power as 100% useful, but such is not the case, as we know there is power loss due to coil heat of
Ploss = I^2 Re = 0.049^2*5.78 = 0.0139 watt
leaving Pmec at resonance of
Pmec fc = IE-I^2Re = 0.0552 watt
not 0.0691 watt such as PF of 1 gives.
And agreeing with PF = I Eb = 0.049 * 1.127 = 0.0552
Bottom line.. PF of 1 does not work, nor does using the PF equation above of PF = Re/Ze, as it gives Pmec fc = IE PF = 0.049*1.41*0.201 = 0.0139 watt Which is the power *lost as heat* not *useful* power Power factor is a bugger-bugger here...
So then the question, what is PF in this case? Or perhaps to be called *effective* power factor???
It is PF eff = 1 - Re/Ze = 1 - 5.78/28.78 - 0.799
Which I have used in areas not discussed here yet.
Then, if we use the usual expression for "real" power (in this case *useful* power), with my "effective" PF of 0.779
IE*PF, 0.049*1.41*0.799 = 0.0552
agreeing as it must to be correct, with
Pmec fc = I * EB = 0.049 * 1.127 = 0.0552 or Pmec fc=IE-I^2Re=(0.049*1.41)-(0.049^2*5.78) = 0.0552
Now... above resonance it gets more (as in MORE) complicated due to reactance. I think you were wise to note
"You have used a power factor as a corrective term. As I said, I would have to look at the pf's as you define them to comment on that."
I have been into this for some time, and as I noted earlier, power factor has been a pain in the arse. I think you are correct that I have used power factor in less than it's usual equation form. In fact there appears a contradiction in the texts between its application to motors vs say, an incandescent light (no reactance), and I've not found an analysis that goes into logical detail (to me at least) on power factor in these aspects. They tend to give power *very* short shrift. Do you know of one?
------------------------------------------
Again before responding to your evaluation of my analysis:
You stated:
"when the mechanical load is not purely resistive then the E-RI calculation must take into account the phase relationships- that is what I have been saying. E and I are not in phase.In this particular case there is a phase shift of 14.46 degrees in the current and considering this gives agreement between E-RI and Eb=BlV. "
Concerning this, you have given the following several times, but without detail of the intermediate magnitudes. Please indicate your calculated magnitudes at each stage, as noted. TIA
------------ It is a bugger but the original electrical definition of power factor really goes back to earlier days (about 100 years ago) when it was observed that , unlike DC, EI was not necessarlily power because of reactance. Hence it was called "apparent power" later named Volt-amperes (VA) as opposed to "real power" in watts. Then VA =S =root(P^2 +Q^2) where Q was the "reactive power" in Volt-amperes reactive or VARs. The ratio of the real to apparent power was called the power factor. Since the reactive component is 90 degrees out of phase with the real component we have a right triangle and PF became known in terms of cos (angle between P and S). For a simple impedance with voltage V applied, the current will lead or lag by some angle as determined by R and X - this angle between voltage and current is the power factor angle.
The concept of power factor distinguishes between real and apparent power (i.e taking into accout reactive) It doesn't distinguish between waste and useful power.
At resonance the input power is as you say: Pin= 0.049*1.41 =0.0691watts. The power lost in the coil resistance is 0.0139 watts as you indicate. The power delivered to the mechanical side is 0.0552 watts. No power factor (in the normal electrical sense ) is involved.
This doesn't mean that the 0.0552 watts delivered to the mechanical side is useful anymore than it means that it is waste. That depends on where the mechanical power is going. All one can say is that Pin= Plost +Puseful.
The electrical definition of power factor can be applied to mechanical systems as well Pelect =EI *cos (angle between F and V) Pmech =FV *cos (phase angle between F and V) "Real power" is due to the component of current in phase with voltage or the component of velocity in phase with force. Whether or not this real power is useful is another question.
---------------- This is correct for Ze purely resistive- otherwise it is not. Isn't it easier to simply say Pout =Pin- losses and calculate the losses as you have done - this eliminates the confusion as to use of the term power factor and doesn't involve any more calculation effort as well as being clear as to each step in the process. It will also work at other than resonant frequency. At resonance you have Ze =Re+R =5.78+23 what you have called an effective power factor is simply 1 -Re/(Re+R) Pin = I^2(Re+R) =0.049^2(28.78) =0.0691 watts. I^2(R) =0.049^2(23)=0.0552 watts = 0.0691 (1-5.78/28.78) Isn't it easier to find I^2(23) than to find I^2(28.78) and then (1-5.78/28.78) =23/28.78 That is the power is simply split between Re and R in proportion to their size.
-------
-------------- If there is any reactive involved, the approach to effective pf that you have will not work. It will be necessary to find the rresistive part of Ze (suppose it is 5.88 ohms) and then remove Re to get 0.1 ohms - then determine the power in this 0.1 ohms as the power transferred to the mechanical side. Total power I^2(5.88) -loss (I^2(5.78) =I^2(0.1)
Basically PF in the electrical sense is as I indicated above in a short history of the term. Note that nowhere in dealing with DC devices, is the term used. It is strictly an AC term and arises because voltage and current are not necessarily in phase. A power factor of unity is desirable for several reasons- mainly minimum current and associated losses for a given real power. There should be no contradiction in its application with regard to motors vs incandescent lamps. The latter are at unity PF while motors generally have a lagging power factor because of inductance.
I think that calling it PF is actually making things more complicated and can lead to problems
All I was doing at each stage was converting from a polar (magnitude and angle) form to a rectangular form (real, reactive) for subtraction then converting the result back to polar form. My calculator handles this quite nicely without my having to do the root or arctan business explicitly.
E=1.41 magnitude RI =1.34 magnitude E-RI =0.353 magnitude Calculation with magnitudes only is meaningless. (E-RI=1.41-1.34 =nonsense (unless I and E are in phase or you are dealing with DC) The phase information is absolutely necessary. (Siskind should cover this).
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
In article , snipped-for-privacy@peeshaw.ca says...
----------
I have no disagreement with your last post regarding parameters at resonance. However, I'm sorry but the above will require clarification. 0.1 appears undefined etc etc. Again, sorry but I cannot follow through it.
---------
In any event, I *finally* found a simple and intuitive relationship between E-(I*RE) and BLv, one that drops (BL)^2/Re out as resonance without using wo/w, fc/f etc or their inverse. My objective at the start of the project was to achieve simplicity and intuitiveness throughout. Now with your insight (and thanks to Daestrom re polarity of Eb) I have achieved tracing the energy chain from electrical through the chain to mechanical, and on into acoustic. Several side facets remain to be ironed out but the pieces will no doubt fall in place. I have several simple equations for f3, a term I call volume acceleration that allows most parameters to be predicted in a *very* simplistic manner, the role of back emf defined in a unique and useful way, etc etc. Some procedures that haven't been noted and/or explained before.
------ Please read through the following before commenting on a given aspect as you go through. TIA
The following is where I fear any more useful input to each other may fail. With due respect for you (I built my first speaker system, such as it was, in 1946...) we will have to have power to push the mass. I cannot follow the energy chain and many other related facets from the mechanical into the acoustic with mechanical power of 0.0054 watt only.
------
I see your point of view that the total power is due to Pmec = v^2*Rms = 0.0493^2*2.23 = 0.0054 when the mechanical resistance in the usual sense of the words is considered, i.e. power dissipated into the resistance of the suspensions.
--------
First note that (Bl)^2/Re must be a part of the resistance (damping) as Rms + (Bl)^2/Re for the mechanical power factor of 0.313 we have used to be valid, i.e. PF mec =Rmec/Zmec=Rms+(Bl)^2/Re/Zmec = 11.12/35.53 = 0.313.
-------
From another view, and during steady state operation: Below resonance the applied force must supply power to overcome the spring-like stiffness of the suspensions plus system resistance. Above resonance the applied force must supply power to overcome inertia of the mass plus system resistance. At resonance the applied force must supply power to balance the system resistance (equal the system resistive losses). As I have noted, only as resonance does the mass get a free ride, meaning the energy required to move the mass and compress the suspension "springs" is being tossed back and forth as kinetic and potential energy. Each opposes the other, they are equal, and they balance out, such that *no outside energy is needed to move the mass per se*. To sustain constant amplitude only enough energy is needed to balance out the frictional (resistive) losses, again *none is needed to accelerate or move the mass* Now.. below and above resonance the mass reactance and the spring reactance are not equal. Looking above resonance, the mass reactance becomes greater than the spring reactance. At 227.4 Hz, Xm = 36.15 and spring reactance Xc is only 2.4 for net reactance of 33.75. This means that to suatain constant amplitude the power supply must now apply more power from the source than just that needed to overcome the frictional resistance of the suspensions Rms. At 227.7 Hz the mechanical power needed is
Pmec = v^2*[Rms+(Bl)^2/Re} = 0.027 watt
Note this agrees with Halliday, Morse, Beranek, etc as
Pmec = Fv cos angle = 1.75*0.0493*0.313 = 0.027
and
Pmec - v^2*Rmec = 0.0493^2 * 11.12 = 0.027
Not equal to I Eb = 0.022. We can get into this later, but note subtracting 0.0054 from 0.027 = 0.0216 watt
Then this magnitude of 0.0216 is in fact equal to
Power accelerating the mass
Pmec m = v^2*(Bl)^2/Re = 0.0493^2 * 8.89 = 0.216
Agreeing exactly note, and *not* coincidence, as the average agreement for the 12 drivers on this is 0.67%.
--------
Now at this point, you may not see power as being dissipated into mass "reactance", that's the rub.. However I can visualize pushing on, say a two foot steel ball, and one may call it mass reactance or whatever, but I will feel it's physical and mechanical RESISTANCE to being moved, and I will have to transfer energy, work even :( *INTO IT* and supply a few joules per second of power to get the mother moving :) In other words, I transfer and DISSIPATE energy and power into it, and it must have RESISTANCE for me to do such. -------
Then from Zmec = sqrt Rmec^2 + Xmec^2 we get Rmec = sqrt Zmec^2 - Xmec^2 = 11.12
Pmec - v^2*Rmec = 0.0493^2 * 11.12 = 0.027
------
Same with Hallidays equation (which I submitted) for the damping resistance term b = 11.12, from which Pmec
Pmec - v^2*Rmec = 0.0493^2 * 11.12 = 0.027
------
Using Rms only in damping also gives the magnitude you now note for mechanical impedance of 33.82. Note that your earlier magnitude of 35.5 is conflictive. As you know, we cannot have two magnitudes. I realize we both have changed a thing or two here and there, but I believe your original Zmec of 35.5 using Rms+(Bl)^2/Re was correct.
-------
Working with force:
R = Fnet/v
Then.. analogous to applied force F app
F app = E Bl/Re = 1.41 * 7.17/5.78 = 1.75
Force driving the mass = back retarding force of the mass is
Eb Bl/Re Blv Bl/Re Bl v Bl (BL)^2 R = ---------- = ------------ = ------------ = ------- v v v Re Re
Showing (Bl)^2/Re is a resistance (whether you want to call it mechanical or electrical) that must be included in the total resistive damping such that again, that total mechanical power is 0.027 watt, and not 0.0054 watt
-------------- I am not sure what you mean by the above. 0.1 appears undefined? Is this in reference to something else that I wrote? All quantities above are volts. The arithmetic is straight forward. What I have said is E-ReI and expanded /converted it from one form to another so that equation A=B=C =D which is the final result in polar form. (the = signs are shown)
-------------- I have read what you have said below and I am afraid that we are going through the same dance again. I will comment at each stage where I feel that it is necessary to address that stage.
It is, as you say, the power into the suspension in this case- absolutely true. This is because, as said before, this particular model doesn't include the acoustic load and, because of that, it is imcomplete.
If the acoustic load is included, then the apparent mechanical impedance is modified. In the simplest case, Rms will be replaced by Rms in parallel with an acoustic resistance Rr (Kinsler does this for a series rather than a parallel model but essentially says the same) and there will also be an acoustic mass . In that case there will be an additional power Rr*V^2 which is the acoustic power. The acoustic impedance will affect the mechanical and electrical Z and the total damping.
With acoustic "resistance represented by an equivalent Rr then Pin =EIcos (input power factor) =Re(I^2) +Rms(V^2) +Rr(V^2)
--------------- I have always recognised that (Bl)^2/Re is part of the damping. Both Re and Rms are damping terms and both must be taken into consideration- I have done that. As to the validity of the mechanical power factor - nothing wrong with R/Z giving a power factor and the use of this power factor provided that it's usage is meaningful. Is it in this case? >
-------
------------------------- I disagree with you on the physics of the situation. Three points
1) In steady state, the average power to "accelerate" the mass is 0. Think of a mass as being a capacitor. During part of the cycle energy is put into the capacitor and during another part of the cycle, this energy is returned to the source. Average power (which you get using rms V,F etc) into a mass is 0. If you apply a sinusoidal force to a mass, the velocity will be sinusoidal and 90 degrees out of phase with the force. (pf=0) This should be in any physics text and I am certain that Siskind will have covered this topic for a capacitor or an inductance. Most electrical texts show the difference between real and reactive power by diagrams showing several cases (R only, X only, R and X ) of the waveforms involved. Only the resistive elements involve an average (as opposed to instantaneous power). The whole concept of power factor is based on this fact of life. Masses, springs, inductances and capacitances are energy storage rather than energy dissipating elements. This has been well established in physics for longer than the sum of our life times.
Note that your power factor is a method of looking only at the component of V which is in phase with the force- i.e eliminating the reactive or mass/ spring related part. Z*pf =R
2)The term (Bl)^2 depends on only two things. The electrical resistance Re and the coupling term Bl which depends on the flux density in the air gap and the length of active conductor in the gap. None of these have anything to do with the mass. Change the mass and the (BL)^Re factor wont change.
As I have indicated before - it is simply an artifact of the model which correctly represents the effect of the electrical resistance on the damping of the system. It has more to do with the representation of the voltage source E behind Re as a current source. The actual force being applied to the mechanical system is the coulomb force BlI=E(BL)/Re -V((Bl)^2)/Re This can be found from E-RI =BlV and BlI =(Rms +jXm)V in the absense of acoustic loading as we have been dealing with this.
3)You have also used expressions for Pmec =Fvcos angle and V^Rmec and attributed them to Berenak, etc but what do they define as Rmec, F, etc in their work? Note that these equations of theirs work perfectly well for the case where F is the coulomb force and the resistance is Rms. Simply plugging into the equation with (Bl)^2/Re +Rms doesn't justify or prove anything. i.e. Pmec =FV cos(angle ) =1.665*0.0492*cos(14.4+71.76) =1.665*0.0492*0.066=0.0054 watts Same formula- different answer.
Have you tried going the other way - refer all the mechanical elements to the electrical side. This gets E=RI +ZmotI Zmot =(Bl)^2 /[Rms+jXms] =(7.17)^2/(2.23+j33.75) =((7.17)^2/33.82) @-86.22=1.520 @-86.22 or Zmot =0.10-j1.517 ohms is the mechanical impedance as seen from the electrical side. Then Ze=R+Zmot =5.78 +0.1 +j1.517 =6.07 @-14.46 degrees (you have disputed this but it is based on data you provided and the calculations have been redone several times but bear with me on the numbers) Then I=E/Ze=1.41/6.07 =0.232 A (at +14.46 degrees with respect to E The pf is cos 14.46=0.968 Pin =I^2(Re +Rmot ) where Rmot =0.1 ohms Pin =0.232)^2(5.78 +0.1) =0.312 +0.0054 WATTS The first term is obviously I^2Re loss and the second is the power dissipated in Rmot Note that I Zmot =Eb so Eb= (1.52 @ -86.22)(0.232 @+14.46) =(1.52*0.232) @-71.76 volts or Eb =0.353 volts magnitude and phase -71.76 degrees V=Eb/Bl =0.0492 m/s at phase -71.76 degrees Sound familiar? F=BlI =1.665 magnitude at phase 14.46 degrees Note BlE/Re -{(Bl)^2}= 1.75 -8.89*0.0492 @-71.76 =1.75 -(0.1369-j0.0.415) =1.613 +j0.415 =1.665 @14.46 This is (Bl)I Also note that the mechanical power provided by the electrical system is given by FVcos (angle) (=1.665*0.0492*cos(14.46+71.76) =0.0054 watts . It IS the power dissipated in Rms as the model discussed so far doesn't include the acoustic load.
The above approach is that used by Kinsler et al. The point to remember is that the results should be the same by using all mechnical analogues as with all electrical or all acoustic analogues.
Note that this is the same as Eb*I*cos (angle) because Eb*I =V*F by virtue of "conservation of energy" which predates the whole concept of a loudspeaker
---------------- You put energy into you big fat steel ball and accelerate it - good- you now have stored kinetic energy in the moving ball. It is not dissipated- just stored. You are experiencing the effect of inertia, not resistance. Friction which exerts a force to slow the rolling ball removes this energy and dissipates it as heat. You can keep the ball rolling at the same speed (in steady state)by only putting in the energy needed to overcome friction. In an oscillatory situation, the same thing occurs - in steady state all that you have to do is supply the losses. In part of the cycle you accelerate the mass and in part of the cycle it decelerates, and then you accelerate it the other way, after which it decelerates again. What energy you put in in part of the cycle is recovered in another part of the cycle.
Halliday said Pmec=V^2*Rmec but he didn't say 0.0493^2*11.12=0.027 I sure hope that his Rmec was Rms (+possible acoustic R) Pmec=0.0493^2 *2.23 =0.0054 also fits Halliday's equation.
(Bl)^2/Re is a useful term but the power V^2*(Bl)^2/Re is not an actual or physically present power. It is meaningless. Using it is like trying to use a Thevenin or Norton source equivalent to determine what goes on inside the source. It doesn't work. I hope that Halliday etc haven;t fallen into this trap. Kinsler hasn't.
---------------- As I said, I did not use Rms only for damping. The (Bl)^2/Re term is the part of the damping that is due to electrical resistance. I used it where it was applicable but not where it is not applicable. >
----------------- Stop right there. You have said that the retarding force of the mass is given by (Bl)^2/Re *V and then did the following calculations on the basis of this premise. That proves nothing as the calcuations are simply different ways to re-iterate the same thing. You have used a circular arguement on the basis that A leads to b which leads to c which leads back to A showing that you have doen your arithmetic correctly but not showing that your original premise or the use of the equations have any physical meaning.
What you have left out is the basis of your premise that the retarding force due to the mass is given by (Bl)^2V/Re ? Shouldn't mass be a factor in there as the other terms except for V are independent of mass. As I indicated above, and physics texts also indicate, the premise is wrong.
--------------- It hasn't. The effect on damping is there. That is true. We agree on that -why keep trying to tell me what I always knew? The rest is where we disagree and I consider that you are wrong. You have not presented anything other than a circular argument to try to show your point. I simply want you to look at the origin of the (Bl)^2/Re term and recognise that power in this is not an actual mechanical power. We have gone over this before repeatedly and the source of the term is very apparent. It is not due to mass, it is not an actual mechanical resistance and the losses in the (Bl)^2/Re are an artifact of the model. In using the equations, you have ignored physical reality. That leads to trouble. Sorry.
At this time there is probably no point in going on. While I respect your experience, remember that I have had about as long an experience, in an environment where everything can be and is questioned, in circuit analysis and electromechanical machine analysis.
-----------
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
You do not currently use (Bl)^2/Re in determining mechanical impedance. Halliday etc do. See below. You originally had Zmec at 35.5 by including (Bl)^2/Re, later changing to 33.82 without (Bl)^2/Re to help support your erreoneous 0.0054 power magnitude.
Yes, since cos angle = PF is used to find the component of force in the direction of displacement. This is covered in most any physics textbook.
(pf=0) This should be in any
You misapply the rule. In either a purley inductive load or a purley capacitive load, the average power is zero, but that is not the case. Nor do we have a purly mass or a purly compliant load. The zero average for voltage and current reflects the characteristic of a sine function (or cosine), but that doesn't mean energy is not being transferred.
P = I^2R
Even though voltage and current change sign each half-cycle, the square of the current is always positive.
Fitzgerald should cover this.
Kinsler covers this with eq. 14.26 R electrical = (Bl)^2 / R mechanical
Same song, different tune. Halliday covers this (see below) with the damping term = 11.12, not just Rms = 2.23. Again... you had it right at the start, until you changed your tune :)
Absolutely not. Net force = force applied * cos angle
F net = E (Bl)/Re * cos angle = 1.75 * 0.313 = 0.548
where angle = arc sin Xmec/Zmec = arc sin 33.75/35.53 = 71.78
where Halliday gives Zmec as 35.53 (see below), and we agree on Xmec. Mechanical power then is
Pmec = Fnet^2/Rmec = 0.548^2/11.12 = 0.027 watt
Same as Halliday and Morse, where
P = Fv cos angle = 1.75 * 0.0493 * 0.313 = 0.027 watt
- good- you now
I suggest you carefully consider this:
"All you have to do is supply the losses" pertains at resonance only, and also with constant velocity in one-direction straight-line motion. This is where I believe you are basically in error on understanding the concept of forced oscillations. Note that both velocity and acceleration are constantly changing and mass *cannot accelerate itself*, although at resonance stiffness reactance does the job.
In part of the cycle you accelerate the
Halliday says it all re our basic difference. See below.
Yes, your usage and non-usage of (Bl)^2/Re gives two magnitudes for mechanical impedance. Which do you want to use today? :)
------------------------
OK. I gave this before, took you several days to respond with how to skirt around it.
Halliday (and Serway) state regarding forced oscillations:
F max x max = --------------------------------------------- sqrt [ m^2 (w^2-wo^2)^2 + (b^2*w^2) ]
We see the damping term is not just Rms, but instead must include (Bl)^2/Re, as I have noted all along.
I believe you now say you agree, if so then you completely misunderstand the damping term's relation to mechanical power, which is at 227.4 Hz
P mec = v^2*[Rms+(Bl)^2/Re] = 0.0493^2*11.12 = 0.027 watt
and not your
Pmec = 1.664 * 0.066 * 0.0492 = 0.0054 watt
-----------
Short cut to the bottom line:
Your analysis gives the wrong magnitude for mechanical power and the wrong (currently) mechanical impedance.
Again you disagree with:
Halliday and Morse:
P = F * v * cos angle = 1.75*0.0493*0.313 = 0.027
Beranek and Villchur:
P = v^2 * Rmec = 0.0493^2*11.12 = 0.027
Colloms:
P = F^2/Zmec^2 * Rmec = 1.75^2/35.5^2*11.12 = 0.027
----------------------------
I agree. You fail to acknowledge your basic errors re mechanical power, angles, and impedance, and dwell on the side issue of the nature of (Bl)^2/Re, which is of no importance re the fundamental equations, i.e. above you state:
"I simply want you to look at the origin of the (Bl)^2/Re term and recognise that power in this is not an actual mechanical power."
The bottom line is:
Clearly, the equations for mechanical power as given by Halliday et al above DO NOT CARE if (Bl)^2/Re is mechanical or electrical. All that matters is that it's part of the resistive damping, giving Pmec at 227.4 Hz = 0.027 watt
Not 0.0054 as you claim. You use the nature of (Bl)^2/Re as a red Herring to divert from your errors.
I am top posting this deliberately, after reading through what you have written.
First of all I gave a long answer to something which can be handled shortly- sorry about that
You spoke of a force in the mass and the power needed to accelerate the mass. The force equation F=[R +j(wM-K/w)]V can be written as F=Fr +Fm +Fk where Fr =RV (disregarding the make up of R) Note that Fr is in phase with R pf =1 Fk ={-jK/w}V and in particular Fm ={jMw}V The magnitude of Fm =Mw =Ma where a is the acceleration. For practical purposes the term "j" implies a 90 degree phase shift and Fm leads V by 90 degrees. Pinto mass =FmV cos (angle) leads to P=FmV (0) =0 As to the angle, look at it this way. - when F is a maximum, the acceleration is a maximum and the rate of change of velocity is a maximum (a=dv/dt) for a sinusoid, this will occur when the velocity is 0 (steepest slope). When the velocity is a maximum, its rate of change is 0 - i.e. it is at a peak ) so acceleration and velocity are 0 This relationship coincides with a 90 degree phase shift. Net result in any case is that the average power into the mass is 0. This can also be shown by going back to first principles and looking at the instantaneous power and determining the average -still gives 0.
------
I looked at what you have written below and it shows a complete failure to understand the circuit analysis or the physical significance of the model and terms in it. You have parroted, ad nauseum , equations due to Halliday and plugged your numbers into these equations as "proof" ignoring the fact that I can plug other numbers into them and get my results. They are not "proof" of anything. You assume that Halliday's Zmec is the same as yours but you have given nothing to show that this is indeed what Halliday meant. If Halliday,etc. use, as you say, a Pmech based on Rms +(Bl)^2/Re to find mechanical power, then they have fallen into a trap that sophomore electrical engineering students have learned to avoid (otherwise they will unlikely be juniors) and are wrong. I don't believe that they are wrong but I do believe that you have taken the equations without any depth of understanding of what they mean and applied them. This is apparent from your responses which often ignore common sense and established physics or circuit theory. This also explains you attempt to tie Re into some relationship with the mass and "power" to accelerate the mass. You also have not had the courtesy of even trying to understand what I have said and repeatedly discredit me with changing of my mind about the mechanical impedance and with regard to damping. YES Re does affect damping and its effect may be represented as an equivalent mechanical resistance BUT V^2 (Bl)^2/Re does NOT represent an actual mechanical power. Please note that I have always included the effect of Re in determining the velocity. I did this right from the start even though you appear to have conveniently ignored that fact. However, I have not used it it where its application is erroneous. That is the part that bothers you. It really is simply a matter of the equivalent and convenient Norton source used in the model. The nature of this equivalent is the "trap" I spoke of above.
For the model that I have used and for Kinsler's model -(see figure 14.10a and the equation 14.51b- not the form of the model that you have referred to- which is valid but a different way of looking at it). are independently obtained models which agree in the end results -The Kinsler approach as an alternative as was worked out below.
Two independent models which agree. If I am wrong- so are Kinsler and his co-authors.
I can look at the other models that Kinsler gave (i.e. the one you refer to) but I like the Zmot model or the model that I have used as being somewhat easier to use.
All that I have written is based strictly on the following: No equations from elsewhere but simply a step by step analysis that I have presented earlier.
Values defined as E =applied voltage, Eb = back voltage, I=coil current, Re =coil resistance, F=total mechanical applied force Zm = mechanical impedance, Rms =mechanical resitance, M =mechanical mass, K =mechanical spring stiffness, V is the mechanical velocity and w is the radian frequency. Bl is the coupling factor relating force to current and Eb to velocity
E-ReI =Eb F=ZmV where Zm = Rms +j(wM-K/w)
Note that neither Eb or F are specified in the above. F can be due to an electrical origin, someone shouting at the speaker cone, fornicating mice, or a combination of these. The force- velocity relationship couldn't give a damn as to the source of the force. It is just a force acting on the actual mechanical impedance. In the electrical equation Eb could be due to an active source , a load impedance - whatever,- again unspecified.
So far, no speaker, just two unrelated equations. (and too many unknowns)
The relation in the actual device is given by the coupling equations
F=BlI and Eb =BlV
Now we can model the speaker from E-ReI =BlV F=(Bl)I =ZmV
Using these values in the electrical and mechanical equations couples them so that the mechanical side affects the electrical side and vice versa.
Do you have a problem with the above? That is all I want to know at this time.
Re will affect damping but, no matter whether you call (Bl)^2/Re a mechanical equivalent - it is still an electrical element and the loss in this element is still I^2Re which is not (V^2)(Bl^2)/Re or the equivalent. This is what you appear to to be unwilling to realise.
As for what you have written below, in the light of my views, above, I will not bother responding to a parroting of the "same old" without the effort to understand what I have said or to really understand why and how the expressions that you quote arise, or to check your results against common sense.Again, I included right from the start, (Bl)^2/Re as part of the damping. It simply falls out of the equations above. Have you used your values of F , V and worked back to find the current and the electrical power input. Do you have discrepancies? If so- why? Have you used Kinsler's approach as indicated using Fig14.10a and Zmot as defined in
14.51b? I have done so below if you would care to compare.
If you don't wish to respond, that's fine too. As I said, I am quite comfortable with my analysis. Results are consistent and agree with an independent model used by Kinsler. At least I know what I have done at each step in the process and have laid these steps out for you in the past.
Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...
I do not consider your top-posting inapropriate, in fact I consider such appropriate at this point.
I'm sorry you feel frustrated to the point of tossing out insults. I see it this way, you may be right or I may be right. So what? The world will go on. Look at it this way.. maybe we'll come up with something worthwhile. After all, the main textbooks avoid what we have discussed like the plague. If so, then one of us losing the "argument" would be a rather small price to pay.
You mention below not responding to reiteration. Tell me about it... Anyway, I agree, and see only *one* need at this time, which is to establish if my premise is correct or incorrect. I'll now state my premise as concisely as possible:
With reference to a loudspeaker motor operating above resonance in a closed box, in this case at the resistive point of 227.4 Hz with my reference driver, and at steady state (constant amplitude) operation, and with constant voltage inptut, I believe this requires that mechanical power be dissipated due to accelerating the mass load, overcoming the resistance of the suspensions, and overcoming the resistance of the air load. Otherwise stated as more power being required than just that dissipated in the suspension resistance and air load resistance.
Also that (Bl)^2/Re or Rme is part of the damping, and that whether considered mechanical or electrical in nature, and since it is part of the resistive damping, it may be included in the resistive term that determines mechanical power, so that average mechanical power Pmec of the reference driver is: _ _ Pmec = v^2 * [Rms + (Bl)^2/Re] = 0.027 watt
corresponding to convention where
_ _ _ Pmec = F * v cos angle = 0.027 watt
Further note that this does not equal a power magnitude of Pmec = v^2 Rms = 0.0054 watt, as stated elsewhere, but includes the power dissipated to accelerate the mass load.
where: _ v = average velocity = 0.0493 Rms = suspension mechanical resistance = 2.23 (Bl)^2/Re = resistance of load due to accelerating it's mass as reflected back through the motor's back emf = 8.89 (Bl) = motor force factor = 7.17 Re = DC resistance of coil = 5.78 F = E (Bl)/Re = total applied average force = 1.75 cos angle = 0.313 E = 1.41
Note that the air load resistance as was measured in this case is included in the Rms magnitude of 2.23. and is Rmr = 0.629
Comments by other than Mr. Kelly are welcome.
Bill W
----------------------
----------------------
----------------------
Again, I included right from the start, (Bl)^2/Re as part of the
"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...
------------------- Actually most texts cover this quite well. It may be that a lot of acoustical engineers are not really at home with electrical circuits and analogs. It also helps to know where to look and Thevenin/ norton models are a good start. I did not intend to insult but it is a fact that some of your viewpoints do show a weakness in the area of circuit analysis and energy conversion principles.
-----------
------------------- I disagree here, and my viewpoint is based on the physics of the situation. This appears in the force equation as the wMV term. The force wmV =MA where A is the acceleration (i.e. F=ma) Note that this force is part of the reactive component of force. As such as explained below, this particular component of the total force is 90degrees out of phase with the velocity. The power factor of this component is 0. No average power input. No energy dissipated in the mass. There is no other term involving mass. As said before, Mass. Spring compliance, Inductance , capacitance are energy storage elements. An electrical analog of Mass is either inductance or capacitance depending on the model. There is no dissipation of mechanical power in driving the mass. Energy put into the mass during part of the cycle is returned to the circuit during another part of the cycle. Net energy =0. I did discuss this previously.
If you have a velocity of 10sin(wt) the force driving a mass of 1kg will be
10*1*w cos(wt) Then the instantaneous power is given by fv =10sin(wt)*10cos(wt) =100 sin(wt)cos(wt) =50 sin(2wt) This is a double frequency sinusoid which has an average of 0. For a resistor F=RV where F and V are in phase. and for a 1 Ns/m mechanical resistance fv=(10*1)sin wt *10sin(wt) =100sin^2(wt) =50 (1-cos(2wt)) which has a non-zero average of 50 For rms values the power factor will be 1 For a general impedance with a phase angle there will be both a constant term and double frequency terms in the instantaneous and the constant term will depend on cos(phase angle)
Now, in the loudspeaker - in the general case the acoustic power will appear as a mechanical resistance. Kinsler shows the simplest case where the acoustic resistance + the suspension damping are combined (Rr +Rmo). Energy is dissipated only in resistive elements (where the V and F or the E and I are in phase) . It's not what you want to hear but it is a fact of life.
------------
-------------- Again, as I have pointed out the term (Bl)^2/Re has nothing to do with the mass . It depends only on the electrical resistance and Bl . There is no relationship between these terms and the mass so it cannot be related to the erroneously claimed power to accelerate the mass. The source of this term is in the nature of the model. If you look at Kinsler's model, where all is referred to the electrical side, this term is not present- but the actual ReI term is present. You had brought up Kinsler's model as in Fig 10b and eq 14.51a (or fig 14.25 and eq 14.26 etc) -note that the electrical terms representing the mechanical system do not include any Re factor. I also note that this model can be found directly from the model that I have used (and apparently used by Halliday etc). I did use Kinsler's model and the results obtained are identical to the results that I obtained. As indicated, if I am wrong then so is Kinsler. You use the equation Pmec =V^2[Rms +(Bl)^2 /Re] to find a power 0.027 watts You also use F*V*pf I use Pmec =V^2(Rms) = 0.0054 watts delivered to the mechanical system (Rms) I get the same using F*V*pf where I use F=BlI The same equations are used. different F or R used. The concept of the "power" into the (Bl)^2/Re term as bing power to accelerate the mass makes no sense as it is simply not related to the mass. The concept that this term is somehow an actual mechanical resistance also makes no sense. The origin of the (Bl)^2/ term has been explained to you before. It is an artifact of the model and while it affects damping, it takes no part in the mechanical power. The model as shown in Kinsler, with everything related to the electrical side doesn't have this term. I suggested that you check up Norton equivalent circuits as that is where this arises. I also asked for your yeah or nay on the basic equations for the speaker before going over this territory again. Anything that I (or Kinsler) have done are based on these equations. and the solutions obtained satisfy these equations.
---------------- Here is a problem: The solutions that both of us obtained for v agree and are based on E=1.41V rms so the velocity obtained is an rms value of 0.0493 m/s - not an "average" value. If it is actually 0.0493 average then the parameters that you have given and listed below are wrong. Which is it?
-----------------------
-------------- Problem- this doesn't have anything to do with the mass and this has been proven before - where the reason that this term appears is given. Look at the development of the equations.
-----------
------------- Again you say "average" force. If the voltage E is rms then the F =1.75 is an rms force. You have used it that way yourself. Which is it? rms or average. Which does your meter measure on the AC ranges? Note that if the velocity is an average velocity, then P=(V^2)*Rmec or P=I^2R are wrong by a bugger factor. (about 23%)
This total force actually exists only at standstill. The actual applied force is BlI =E(Bl)/Re- V (Bl)^2/Re where phase angles must be taken into account in the calcuation. This was shown. The (Bl)^2/Re term as well as the F=E(Bl)/Re are a result of the source equivalent used See Norton and Thevenin equivalents.
my position:
1) (Bl)^2/Re is an equivalent mechanical impedance representing the effect of Re on the damping i a model referred to the mechanical side. It is not an actual mechanical impedance and power V^2((BL)^2/Re is meaningless in that sense. I am arguing from both circuit analysis techniques (which are used here) and from simply recognising that Re is an electrical element
2) the attempt to relate the so called power above to some power "dissipated" in the mass is a severe violation of mechanical principles (as well as circuit principles). In any case this term has no relationship to the mass.
3) we must get it straight- either the values being calculated are average or they are rms. If E is rms volts then the calculated V, I , F etc are rms not average. This appears to be the case. If the F and V are average and E,I are rms then Bl is is error. If V is rms, then the max V that you gave is in error. If F,V are average values then the expression you use for power is incorrect.
-------------
---------- Thank you.
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
In article , snipped-for-privacy@peeshaw.ca says...
Why do you continually rant about my "weakness", etc etc? I am beginning to think you are here just to jab at someone. Otherwise??? Surely not :) Really... you need to cut the personal crap.
Regarding the basic disagreement, you need to look at yourself instead of me, maybe by reading and making an attempt to understand these textbooks you speak of.
You say below: "Again, as I have pointed out the term (Bl)^2/Re has nothing to do with the mass. The concept of the "power" into the (Bl)^2/Re term as bing power to accelerate the mass makes no sense as it is simply not related to the mass."
You are wrong, as I have a simple equation relating mass, the power needed to accelerate it, and (Bl)^2/Re. Also a simple equation relating back emf (the medium of transfer between the load (mass acceleration) and the applied power. But we'll never get there until you understand and acknowledge that IT TAKES POWER TO ACCELERATE THE MASS *****ABOVE RESONANCE*****, plus the load due to acceleration is reflected back through the (BL)^2/Re resistance. I have shown this and you did not comment, skipping over it and on into another dissertation with your math and math and math. I have also told you I have an equation that drops (Bl)^2/Re out at resonance, I believe this is your hangup, i,e. that you cannot use (Bl)^2/Re at resonance in the power equation. Of course you can't, the mass exhibits no power requirement at resonance to be accelerated. I have tried every way I know how to get you to see that power IS required above resonance to accelerate the mass. Perhaps this will help: (Bl)^2/Re is not the intuitive expression for the resistance due to accelerating the mass. (Bl)^2/Re is derived from the back resistance equation. I gave you this. You skipped over it as well, commenting something about my "ignorance" :( Bless you and the horse you rode in on, but you must knuckle down and address the above. Not another multi-page math speil. Just address the above please. Note... just the above please.
Then ABOVE RESONANCE:
Pmec = v^2 * [Rms + (Bl)^2/Re] = 0.027 watt
------
You say below: "the attempt to relate the so called power above to some power dissipated" in the mass is a severe violation of mechanical principles (as well as circuit principles). "
Are you resorting to dishonesty? Please quote where I said power is dissipated in the mass. I want to see that. I said power is required to ACCELERATE THE MASS above resonance.
"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...
--------- It is not personal. It is an observation based on the sort of nonsense that you spout below. I don't know your strengths as you haven't shown them. The observation that they are not in circuit analysis or electromechanical analysis is based on what you have shown. I am not going to agree with what is patently wrong and have given reasons why these concepts are wrong- the fact that you seem unable e to follow what is really sophomoric or (even some grade school stuff) is part, but only part of the basis for this observation.
-------- The trouble is that I do understand them and the math and physical principles used in them.
I thought of answering it but it wasn't worth the hassle-I simply ignored this as it was patently nonsense.. The math I have used is fairly simple and is pertinent. The whole model of the speaker that we are considering is essentially a mathematical one. Math is the tool of choice. It helps to understand the tools rather than 'plug and play". Did you ever consider trying to sketch power vs time for a mass (or a capacitor, a spring or an inductance) I suggest that you do.
Please tell me how Re and Bl are related to mass? You haven't answered that. Since the so called "power" to accelerate mass will depend on frequency (velocity and acceleration) , please note that freqency is not a factor in (Bl)^2/Re.
Yes, with a higher V, the "loss" V^2(Bl)^2/Re increases- the effect that the mass has is that for a given source force the mass will affect the velocity. That is true. It is in the same category as adding a capacitor in parallel with a resistive load will increase line losses. No more than that.
The "load due to acceleration" is there- I have pointed it out and exactly where it appears in the equations. It is given by wMV which is a force but it is a reactive force. since it is 90 degrees out of phase with the velocity the power associated has a zero average unlike the power in a resistive element which has a non-zero value.
Basically you have made a claim but there is nothing to justify either the basic premise or the conclusions obtained. Sorry, it is a non-starter.
Whether you are above resonance, below resonance or at resonance- there is no average power involved in accelerating the mass Zero, Zilch. I have given you an explanation which you proceed to ignore. Also why does mass not appear in the "resistive" part of the impedance and why does the use of power factor eliminate the effect of the reactive part which involves mass. I suggest that you go back to Siskind and really look at power factor in terms of the relative phase angles between voltage and current (or force and velocity) Note that the electrical analogue of mass (as used in this model) is a capacitance - I have also given you references that should help you.
-------------- Oh, but it DOES work just as well as it does at any other frequency. Please note that at resonance E(Bl)/Re =((Bl)^2/Re +Rms)V or 1.75 =11.12V or V =1.75/11.12 =0.157 m/s Remember that number? See the (Bl)^2/Re term?
(Bl)^2/Re doesnt change with frequency - the acceleration of the mass does.
The equations at resonance are the same as at any other frequency-just the numbers differ as affected by frequency. Now using YOUR form of analysis and noting that the power factor is unity (F and v in phase) then the power is Pmec =11.12(0.157)^2 =0.0247 watts (note that (Bl^2) /Re is still there.
This is NONSENSE for the same reason that it is nonsense at any other frequency including 227.4 Hz.
The reason is the same. (Bl)^2/Re is due to the electrical resistance. Converting it (and BlE/Re )to the electrical side will give a current source of E/Re shunted by a resistance Re (Norton equivalent) which can be converted to a voltage source E in series with Re --leading , incidentally, to the equivalent circuit shown by Kinsler. That is also the reason why I suggested that you look up Thevenin and Norton equivalent circuits.
If I use my approach or Kinsler's (which doesn't have the (Bl)^2/Re term- does that mean the mass is not being accelerated?) - they end up with the same results
F=BlI =0.157*2.27 =0.35N, Eb=1.128 V, I =0.049A These lead to Pin=1.41*0.049 =0.0691 watts I^2Re =0.0139 watts and Pin-I^2Re =EbI =FV = the air gap power transferred to the mechanical side =0.055W Using Pmec = V^2 Rms =(0.157^2)*2.23 =0.055 watts. which agrees as it should. I would be quite worried if it didn't. .. Have you tried this?
Please look at the original equations and see how the (Bl)^2/Re and the (Bl)E/Re terms arise. I have asked you to do this and also have shown this. I am willing to go over this again but are you willing to look at it and ask questions about the parts that you may not understand.
I know how (Bl)^2/Re appears- long ago I developed this from the basic equations. I also recognise how it arose from those equations and why any power calculated from this is meaningless.
----------- NO. That is wrong. Sorry. I won't repeat any math or verbal explanation as to why as you have simply ignored the explanations that I have given. Oh, yes you kept repeating a basic equation, disregarding the fact that I could use the same equation with my numbers to get my result. You also ignored the fact that using Kinsler's model which doesn't have the (Bl)^2/Re term, the results are the same as what I got. All I ask is that you do two things- study Siskind to see what happens in an inductance or capacitance and note that the model used is an electrical analogue. In addition Siskind should have something on Norton equivalents and equivaent sources in general. All that is involved in this whole thing is some simple circuit analysis. ----------
--------- I recall the word "dissipation" being used as well. . Be that as it may, my comment stands. Below resonance, the mass is still being accelerated but the spring is dominant. Are you now going to tell me that the V^2(Bl^2)/Re is the power to compress the spring?
There is instantaneous power into the mass during part of the cycle (never denied) but at other parts of the cycle, the power is returned to the supply (the mass decelerates and returns energy then is accelerated in the opposite direction. then decelerates again and the process is repeated). Energyis stored and released repeatedly and power flow shuffles back and forth. Zero average. I gave you an equation for this which you could look at and plot. There is an associated force which is clearly shown in the equations for the device. (fmass =wMV in magnitude but a 90 degree phase shift). This is the only mass related term. If you are saying that the power lost in the resistance is affected by this- that is true because mass will affect the impedance and the force/velocity relationship. If you are saying that it is a measure of the power to accelerate the mass- then that is nonsense. This shuttling of power is called, in the electrical world , volt-amps reactive or Vars. This is the out of phase component of EI. This component increases the current for a given power and voltage hence the whole basis of the concept of power factor. While the the mass does change the velocity (and losses) for a given force, this is shown in its effect on the power factor. Whether you are working with a mass or a capacitor or inductor, the same principles hold. I also note that the model in which (Bl)^2/Re appears is essentially an electrical analog in which force is treated as current , velocity as voltage, mass as a capacitor, etc. Mechanical impedance appears as an electrical admittance (and this is also done by Kinsler).
As a final comment, please note that I have worked it out using the models that Kinsler (and have shown you the results) and note that the results are the same as those I obtained. That seems to corroborate my position. Have you tried to do the same?
I made 3 points. You have certtainly not answered any of them with any sort of logical argument. You have made a premise and a conclusion with no support for either of them in termsof logic or otherwise.
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
Nonsense, eh? We'll see below whose effort is nonsense.
Here is where you show your hand. First, as reference, you give the magnitude of mechanical power at 227.4 Hz as Pmec =V^2(Rms) = 0.0054 watts
However, Rms is the suspension resistance, so this is the power dissipated into the resistance of the suspensions leaving no power to accelerate the mass, which must be done above resonance for constant amplitude motion. My equation for mechanical power at 227.4 Hz is
Now... I say above that it takes power to accelerate the mass above resonance, and you will not address this because you know it is true, and to admit such is to admit your power equation is wrong, as it contains no power to accelerate the mass. You jumped on my "lack of common sense" and got caught in your own trap. Too bad, since it invalidates your analysis as well. So much for hi-level math without common sense.
Now we'll move on to more of your nonsense.
Just waiting out of politeness, to see if you would admit your errors. Here is one relation of mass to (Bl)^2/Re (I have many):
Note (Bl)^2/Re is 8.89 so this is within 0.3%. The twelve drivers average slightly over 1%. Should be close enough for you. Note I gave magnitudes, unlike you who doesn't return the favor. Enough said.
Ah ha... you spill the beans explicitely. Of course it takes power to accelerate the mass above resonance. This statement along with your power equation, says it all and invalidates your analysis. BTW, I suppose you think those amorous mice you mentioned earlier accelerate the mass at 227.4 Hz. This is now ludicrous, and beyond hope.
It ain't me that needs help here.
1.75=11.12V is absolutely incorrect without considering phase, which you fail to indicate. You may need a break.
I see why you don't give many magnitudes. You say here:
Pmec =11.12(0.157)^2 =0.0247 watts This is wrong. Pmec =11.12(0.157)^2 = 0.274 not 0.0247.
Also power *input* at resonance is only
Pin @fc = IE = 1.41 * 0.049 = 0.0691
So by your calculation, you have more power out than power in. Also the conventional equation for power out in all textbooks of repute that I know of gives mechanical power out of
Pmec = I Eb = 0.049 * 1.127 = 0.0552
Did you shift the digits (2&7&4) trying to avoid violating the conservation of energy, hoping the math would not be checked? In any event it appears you need a break.
Well, we saw nonsense here, alright...
I suggest you look at the back resistance of the load expression. Call it Rb, and Fb is back force due to the *MASS ONLY* of the load. I gave it before, look at it this time. Check it out, it's simple enough. At 227.4 Hz
Eb (Bl)/Re (BLv) (Bl)/Re (Bl)^2 Rb = Fb/v = ------------ = --------------- = ------- = 8.89 v v Re
Look familiar, as you say? Had to wait for it a second time, didn't you? Now.................................
Note it is a resistance dependent on the back force of the load and velocity, and is makes itself known in magnitude through the medium of the back emf. Remember that back emf is noted in the textbooks as linked in magnitude to load-size and velocity? I have a feeling I'm "talking to the wind" here...
Anyway... Again, then:
Pmec = v^2 * [Rms + (Bl)^2/Re] = 0.027 watt
See (Bl)^2/Re in there? It belongs in there as a dissipative element. Energy to push the mass. Power needed to push a few joules per second. To the devil with the armorous mice, never cared much for the little rascals anyway.
OK, you base your argument on this, and as you stated earlier "if I am wrong then so is Kinsler"
This is easily dispensed with:
Note that I use (Bl)^2/Re above resonance at ***227.4*** Hz. Please prove (page number will suffice) where he says your contention applys above resonance. Failing this, I shall consider your argument moot.
Really, my best suggestion is for you to obtain a copy of Beranek's Acoustics. He laid it in non-frequency specific terms in eq. 7.1, where v = F/Z with these parameters broken down into all appropriate components of each (phasors not needed, note). To my knowledge, the equation has never been disputed, and it is the basis of much analysis.
Checking this equation and it's import out at the outset could have saved you a lot of grief and embarassment. However, it appears you were more interested in zapping me, and you fell into your own trap. Too bad.
As I have noted ad nausem or whatever, (Bl)^2/Re drops out of my power equation at resonance, since mass requires no power at resonance to sustain constant amplitude of motion. This has become inane, or worse.
See the above back resistance equation.
Well... lets not say I ignored them, but the above shows why I didn't accept them.
Do you even realize that things are different at resonance from above resonance regarding what we discussed? Get a basic electronics guide, for goodness sake!
So.. you claim I said power is dissipated in the mass, quote: "the attempt to relate the so called power above to some power dissipated" in the mass is a severe violation of mechanical principles (as well as circuit principles). "
Then you cannot prove it, and say your comment stands. There is no point in my continuing to take your communication as anything but tripe, and using the word tripe is extremely benevolent here.
See above re Kinsler and frequency specific equations.
Like I said, best not to take your word. Support for my contention was offered at least twice. One last time, here it is again, in case newcomers are reading this:
We see the damping term is not just Rms as in your power equation, but instead must include (Bl)^2/Re, as I have noted all along.
Giving mechanical power at 227.4 Hz of
P mec = v^2*[Rms+(Bl)^2/Re] = 0.0493^2*11.12 = 0.027 watt
and *NOT* your magnitude of
Pmec = 1.664 * 0.066 * 0.0492 = 0.0054 watt
----------
Halliday, Morse, Beranek, Villchur, and Colloms prove you wrong:
Per Halliday and Morse:
P = F * v * cos angle = 1.75*0.0493*0.313 = 0.027
Per Beranek and Villchur:
P = v^2 * Rmec = 0.0493^2*11.12 = 0.027
Per Colloms:
P = F^2/Zmec^2 * Rmec = 1.75^2/35.5^2*11.12 = 0.027
Note that all agree exactly, giving magnitude for net mechanical power of 0.027, same as my magnitude, and not a magnitude of 0.0054 as you claim.
You're up against a tough crowd here, and out on a limb all by yourself. Time to climb down and join the crowd.
Anyway... if you still disagree with the authorities above, then that will speak for itself, and obviously you will have problems with the subject at hand at least, and beyond anything more I can help with.
Perhaps I'll upload a post later with my view summarized for comments separate from your input. Anyone wanting me to do such for discussion may email me at:
snipped-for-privacy@mounet.com Subtract 1 from 5 for email.
----------- "Real" power is not necessarily useful power. power losses are also "real" power. Real power is the rate of energy transfer or conversion.
EI is "apparent power" (old terminology) in units of volt-amperes (VA) EI*pf is the"real" power (component of I in phase with E) (Watts) EIsin (angle between E and I) ="reactive power "(again an old term) referrred to as Volt-amperes-reactive or VAR pf will be lagging if current lags voltage. In this case the Vars are considered + Pf will be leading if current leads voltage Vars are considered negative Think of a right triangle (power triangle) base =P in watts, opposite side =Q in VAR and hypotenuse S in VA Actual power is the base quantity P=S* cos (angle)
At resonance, the voltage and current (as well as force and voltage) will be in phase- 0 angle so the power factor will be cos(0)=1 or unity.
Efficiency is a different story
n=Pout/Pin =(Pin-Plost)/ Pin =Pout/(pout-Plost) Pout, Plost and Pin are all "real" powers Your efficiency is correct if the IEb is all useful power. This is the electromechanical efficiency (power transferred to the mec side /electrical power input)- not the overall electroacoustic efficiency. Your Pf fc is actually this electrical to mechanical efficiency as you have realised.
Since it is nice to have relationships that work at any frequency it is also nice to keep the concept of power factor and efficiency separate.
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
---------------- Lets look at this premise - first of all, you have given no analysis that supports this stand (what you have given below doesn't do this -more later) If you had bothered to do as I suggesnted- that is look at the instantaieous force and velocity through the cycle and calculated the instantaneaous power from p(t)=f(t)*v(t), what I (and textbooks around the world) do say with regard to reactive elements such as mass, springs, inductance and capacitance. Here is a step by step approach which avoids the math you don't like.
1)Sketch velocity v(t) vs time -i.e. a sinusoidal function v(t)=sin wt where T is the period. Pretty standard sime wave graph. This will be 0 at wt= 0, wt=Pi and wt=2*Pi. It will be a positive maximum at wt=Pi/2 and minimum at wt =3Pi/2 (or a negative maximum if you prefer).
2) Note that the rate of change velocity with time is positive between wt=0 and wt=Pi/2 and also that this rate of change is maximum at wt=0 and decreases to 0 at wt=Pi/2. The rate of change of velocity is negative (velocity decreasing) between wt=Pi/2 and wt=Pi.
3) the acceleration at any instant is given by a=rate of change of velocity at that instant. The acceleration and the related f(t) =Ma(t) is positive between wt=0 and wt =Pi/2. It is a maximum at wt=0 and is 0 at wt=Pi/2. . Between wt=Pi/2 and wt=Pi the velocity is decreasing so the acceleration and force are negative.
4) The situation for wt=Pi to wt=2*Pi is similar (from Pi to 3Pi/2 the velocity and the acceleration are both negative. From 3Pi/2 to 2Pi the velocity is negative and the force is positive. Overall, the force (or acceleration) plot is sinusoidal but shifted 90 degrees so that the positive maximum is at wt=0. (F leads V by 90 degrees) The chart below summarises the situation wt v f p=fv
0 to Pi/2 + + + Pi/2 to Pi + - - Pi to 3Pi/2 - - +
3Pi/2 to 2Pi - + - At wt =0, Pi/2, Pi, 3Pi/2 and 2Pi the power =0. power is delivered to the mass in the periods 0 to Pi/2 and Pi to 3Pi/4 Energy is being stored in the mass as KE Power is reversed in the periods from Pi/2 to Pi and 3Pi/2 to 2Pi - stored KE is being returned to the circuit. The average power per cycle is 0 as the acceleration periods are balanced by deceleration periods.
Why not do the above exercise and see for yourself. There is a shuttling of energy into the mass and out from it and this occurs twice per cycle. Only in the case of a purely resistive element, will the power always be positive. In any R-C or R-L or Rm-M device there will be a period of time where the instantaneous power reverses - .
------------------------------ This is cute. It is a case of plugging in numbers to get a magnitude without looking at what's behind the numbers or defining the terms used. Please note that the frequency response and the half power points are dependent on the physical and independent parameters Re, Rms, M, K and Bl . The Q factors are defined in terms of these. (you have conveniently not given the definition of Qmc and Qec). Also note that the spring compliance can be eliminated by use of wc^2=K/M . If you care to go back to the definitions of the various Q's and the way they can be derived from the basic equations you will find that all you have done is written an equation that boils down to (Bl)^2/Re =(Bl)^2/Re. Is this a startling revelation? If you change M in this equation, (Bl)^2/Re will not change but F1,F2, Qec and Qmc will change. Sorry- this is not supporting your contention. You have failed to show how the electrical resistance of the coil, the strength of the magnetic field or the length of conductor in the magnetic field are affected in any way by the mass. I am still waiting for this.
You want to work with numerical magnitudes only so you can't see the relationships involved. Why not do some algebraic work so that you can actually trace the terms and what happens to them. Also, I gave you all the magnitudes that were needed and then some- so you can get off that kick.
----------------------
------------ How so? Please note that wMV is a reactive term- V and wMV are 90 degrees out of phase. I have clearly stated this and it definitely is so in both your and my formulations for Z.
Purely reactive - no average power. (see above) Open your mind - try thinking. period of acceleration =period of deceleration. power in for part of the cycle, power out for part of the cycle- average is 0.
----------------------snip-------------
------------ Since the phase angle is 0 for all quantities as I have indicated previously , the above is absolutely correct. Please note that in the above Z=11.12 +j(Mw-K/w) =11.12
+j0 == 11.12 +j0 =11.12 @ angle
The rest follows I was giving you credit for the ability to at least realise that.
--------------
--------- Typo. I stand corrected
----------------
------------ The digit shift (and factor if 10 was a typo but the corrected results worsen your case. Please read what you have said. a) Pmec =0.274 b)Pmec =IEb =0.055 watts c)Pin =IE=0.0691
b and c are correct and conservation of energy reigns supreme . electromechanical efficiency =0.055/0.0691=0.80 ( to 2 decimal places as the current value used was good only to 2 decimal places. ) I believe you correctly got
0.788 in another communication).
However, taking your value of Pmec (which makes the situation even worse than my typo had done) Pmec =0.274/0.0691 =4.0 (to 2 decimal places )or an electromechanical efficiency of 400% - perpetual motion machine.! I admit to a typo and not checking this but all your correction has done is to reinforce my point. Thank you.
-----------------
incidentally,
------------ This is very nice but it is garbage.
1)You have ,without any reason or rationale said that Fb is the back force due to the mass only- no reasoning given except "I said so"
2)Your equation that Fb/v =(Bl)^2/Re is correct but it simply doesn't imply that this is due to *mass only* as you assert. It simply expresses the fact that electrical resistance is reflected to the mechanical side as (Bl)^2/Re - no more than that. Re exists, (Bl)^/Re is a circuit elemnt which reflects the existence of Re. It is no more than that. Again, this falls out directly from the original basic equations.
3)I don't think that you have yet, despite Daestrom's efforts, got a really solid understanding of back emf. It is simply, just as stated by Eb=BlV, nothing more than a voltage induced by the motion of the coil. The "size" of the load or the source of the coil motion are not involved per se. It is nothing more than a velocity induced voltage. The linkage appears because there is also the F=BlI term and this, along with Eb=BlV provide the coupling between the electrical and mechanical systems. These reationships go together in the case of an electrical excitation. Please go back to the original equations with or without mice rather than make unsupported claims to try and justify a term that exists due to the coupling of the electrical and mechanical systems. Also note that one could consider a ficticious speaker with no mass, no spring stiffness - just Rms. Then, at ALL frequencies, you will have F=(8.89 +2.23)V just as at resonance The (Bl)^2/Re term exists with no mass. In fact the so called power that is dissipated in (Bl)^2/Re will be 0.22 watts at any frequency. Now add mass and at 227.4Hz, the velocity is lower (
---------------------- . (Bl)^2/Re IS a dissipative element. True. However we are back to the faulty premises that (a) there is an average (as opposed to instantaneous for part of the cycle)power required to drive the mass, and (b) that such power, if it existed, is represented by V^2(Bl)^2/Re. It is obvious that you have not looked at Thevenin or Norton source equivalents- Do you want a summary of these?
--------------- There is no restriction on w and the form of this term is the same above and below resonance. Kinsler doesn't say that the expressions that he has or the models apply above resonance. This is taken for granted. Please note that the circuits and equations on pp359-362 as well as the material beyond that do not assume any frequency limitations. Kinsler gives plots of R and X for which cover a wide range of frequencies. Notably figs 14.13 a to e. 14.13e exends to 10KHz. The model that either you or I have used will not agree with this at higher frequencies because Kinsler uses an acoustic resistance that varies with frequency (p364) and also includes the coil inductance- both factors that we have ignored for simplicity (i.e Rms =constant). Both of these are easy to include. Ignoring them will give a different high frequency response than that which actually occurs.
p368 covers other higher frequency factors which come into play about 10 times the resonant frequency and for which models based on the parameters that you have given are not valid. Theile -Small parameters as measured are not valid in this range.
----------
---------- I have no problem, as I have stated before, with v=F/Z as you have given, for determining the velocity magnitude from the force (BlE/Re) and the inclusion of (Bl)^2/Re In fact, If you would look at what I did, I do use these values. Without seeing the equations and the breakdown, I cannot comment further. I use phasors as this is basically electrical circuit analysis rather than Beranek's area of specialty - acoustics- phasors and various circuit analysis methods are a very convenient and highly developed tool for this. I will try to get Beranek from the person who loaned me Kinsler (he has it in a box somewhere) >
--------- Actually, at resonance you have used my approach. The difference is that I use it at all frequencies because for the model used, it is good at all frequencies. You calculate V =1.75/Z =1.75/11.12 = 0.157 ----> note that the (Bl)^2/Re term is present. ---I do the same. Then you recognise that the "actual" mechanical impedance as seen from the electrical side is given by (Bl)^2/Zm =(7.17)^2/2.23 =23.0 ohms ---- I do the same. The total resistance is then 28.83 ohms and you then essentially calculate the total input power (EI ) =I"power factor " 5.78/28.8=0.80 and multiply it by a "power factor" and some manipulation to really say that I^2(28.78) =(I^2)(23.0)+I^2(5.78) and eventually work this around to say that I^2(23.0) =I^2(28.78(1-5.78/28.78)=0.0691(1-5.78/28.78)=0.055 . This is simply the same as saying that I=1.41/28.78 =0.049A and then the mechanical power =I^2 (23.0) =0.055 Note that the mechanical power depends on (Bl)^2/2.23 which is in line with what I have been saying all along. Except for the awkwardness and the distortion of well defined terms, the approach that you use is essentially the same as that which I have used. Your convoluted approach referred to mechanical side quantities would become, at resonance, Pmec=(0.157^2)*2.23=0.055 which is exactly what I have been saying Pmec=(V^2)*Rms
I have eliminated the Re related term from power calculations at all frequencies while you have eliminated it (correctly but based on incorrect reasoning) at resonance only. I use (BlE/Re)=V[(Bl)^2/Re + Zm] at all frequencies (Zm =Rms +j(wM-K/w)and then evaluate mechanical power at any frequency by using Pmec=RmsV^2 at all frequencies. All that the frequency is doing here is changing the magnitude and phase of Zm. It isn't changing (Bl)^2/Re although the acceleration is frequency dependent. >
--------- Not really The same equations are valid below, at, and above resonance. Look at Z=R+jX where X=(wM-K/w) There is nothing in that to make it different at, below or above resonance. All that changes is the variation of X with frequency which means only that Z at one frequency will not be the same as Z at another frequency and resonance is the condition where Z=R+j0 or wm-K/w =0. No more than that. It is an important special case -no denying that but "different" - that it is not. Literature doesn't deal specifically with "above resonance" conditions as, within the parameters of the model being used, there is no essential difference. It may be that the model has to be modified at higher frequencies but that is not related to "above" or "below" resonance but simply means that other parameters such as inductance must be included in the model. I do have good electrical (as opposed to "electronic") circuit texts and have had and used quite a variety of these over the last 50 odd years (all using complex numbers - Oh my!) and these don't support your supposed difference.
------------
-------------- I agree with the above statement- I lso note that you said, and I quote:
"Now at this point, you may not see power as being dissipated
-------- Which do you want?. Make up your mind
-------------------
------- You are right- it is benevolent.
-----------
_------------ ??? - His equations *with the parameters that you provided* and without the parameters that you have ignored, give the same results that I get. This has been answered above. Right from the beginning, I indicated that there are limitations on the model. However, these limtations have nothing to do with 'below resonance, "above resonance" or "at resonance" but have to do with the inadequacy of the model at high frequency. That is, a better model takes into account the frequency sensitivity of the acoustic impedance and the coil inductance- data that you have not provided nor taken into account.
-----------
------------ Again, deliberately or otherwise, you have ignored what I have said. The damping term does include (Bl^2/Re I have not denied this. The equation that Halliday has given is in full accordance with what I have said. It is correct but your assumption that it gives the actual mechanical power is incorrect. Note that Halliday's expression does not deal with power. Also note that Halliday's expression as I have said before, can be developed directly from the impedance expression Z'=(Bl)^2/Re +Zm and noting that Mw-K/w) =M(w^2-wo^2)/w and F/X =wF/V
-------------------
-------- No it doesn't- even if you say so. It doesn't deal with power. It simply is an expression for the total system damping including both electrical and mechanical damping. The (Bl)^2/Re term reflects the electrical resistance as seen from the mechanical side. The power in this term is meaningless - a result of the model, (you obviously haven't looked up Norton and Theveinin equivalents - Just how old is your copy of Siskind?)
------------
----- OR, using the Lorentz force and the actual mechanical impedance P=1.664*0.0493*cos(86.22)=0.0054 WATTS
---------
--------- OR p=(0.0493)^2 * 2.23 =0.0054 W
----------
------ OR p=[(1.664)^2/(33.82)^2]*2.23 =0.0054
--------- Same equations -different results- the equations don't prove your point. However, the following support my point. Another OR
airgap power is given by EbI cos (angle) =0.363*0.232*cos(14.46+71.76) =0.0054 OR EI cos (angle)-I^2Re = 0.0054 Conservation of energy is satisfied with my values. It doesn't hold for your values. Are you implying that the mechanical power exceeds the power converted from electrical to mechanical power and that your sources would agree with you that the mechanical power out is 5 times the air gap power? This is something that you implied above - at resonance.
---------------
------------- The above expressions, as I have shown work equally well with F=BlI and Rmec=2.23 and satisfy the prime test- conservation of energy - so , as said before, what are you proving by the above? Could it be that I am with the crowd and your interpretation of their equations is wrong. Again, you have plugged in numbers without thinking of the physical facts behind the equations. Your argument and reference to these equations above doesn't justify your position. . I note that you have not defined just what *they* mean by F, Rmec, and Zmec. It certainly is not what Kinsler means.
-------------------
----------- I don't disagree with them -UNLESS you have correctly interpreted what they have actually said (i.e. what they consider as F and - and you have given no indication that you have done so. I do disagree with what you have said
-
----------
Do that- but first do some serious questioning of your view and the discrepancies that arise. I won't be answering publicly with long responses like this, or possibly, at all, unless you do seriously consider some of these points. It really isn't worth the effort for such a simple situation.
Thank you
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.