Motor torque and back emf

Kelly replied: " 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 to follow what is really sophomoric or (even some grade school stuff) is part, but only part of the basis for this observation. "

End of reinstating Kelly's snipping.

----------------------------------------------------

I ignored your insults (an example given above that you snipped)long enough, Kelly. Now you take 3 or 4 days to conjure up another **LONG** and senseless diatribe, in an attempt to avoid admitting your errors. That has to be painful to you, but one usually pays for their mistakes one way or another.

There is no point in attempting to address your endless drivel, when you refute the most basic relationships set down by the experts. All the experts give mechanical power at 227.4 Hz of 0.027 watt, not your magnitude of Pmec = v^2*Rms = 0.0054 watt. For the (third?)time, your equation gives the power dissipated in the suspension resistance only, with no power shown in the equation to accelerate the mass, which is required above resonance for constant amplitude motion. Even the basic work equation, which uses measured magnitudes (***separate from whatever resistance you or I claim***) proves you wrong, where at 227.4 Hz and starting from zero velocity P=W/t = 1/2 m*v^2/t = .5*0.0253*0.0493^2/.0011 = 0.0279 No doubt you will refute this as well. I did learn a thing or two from your math and I appreciate that, but it was not really worth the hassle. It's unfortunate that you were arrogant and condescending almost from the start, bent on showing me up, and fell into your own trap. Worse still, you have dug your hole deeper and deeper in an attempt to avoid admitting your errors. You are not looking good here Kelly, and I'm almost sorry I was instrumental in this coming about.

No further reply is given here to your endless drivel below.

Bill W.

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Reply to
Bill W.
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Dear Bill,

I admit that I did lose patience and wasn't as polite as I should have been. I certainly didn't intend to be condescending but when I question things and repeatedly get non-answers or answers which are not based on logical analysis or on garbage values in a valid equation, I get a bit testy.

You keep claiming that I disagee with the "experts" but you have not shown this- all you have done is plugged numbers into an equation attributing these numbers to the experts. I have no problem with the equations but do have troubles with your misapplication. As far as the acoustics is concerned, I give way to these experts, BUT, as far as the "circuit analysis" that they use, I have as much or more expertise- that has been my business. That is not an attempt to boast. I have suggested a way to look at the power into (and out from) a mass and you have not even given it the time of day. I have pointed out the many discrepancies that your approach leads to and such discrepancies do not seem to matter. The fact that (Bl)^2/Re is independent of mass doesn't seem to be of consequence. you have not provided any answers to these discrepancies. The fact that this term arises out of the basic equations (without even considering the details doesn't seem to be of interest For example, you seem to ignore the fact that your numbers lead to more output than input- a problem that you haven't addressed. Also, it doesn't seem to be strange to you that below or above resonance, as you get closer to resonance, the trends in input power by your calculations point to a power of the order of 0.275 watts no matter how close one gets to resonance etc, from below or above and a sudden change to 0.055 at resonance. Since wM-K/w changes smoothly through resonance, there appears to be a problem with that. Try it.

For example: Below, you give a power into the mass using KE which is based on 1/4 cycle. If you had used v as the peak velocity, it would have been correct for that

1/4 cycle. {stored energy at start =0, at end it is 0.5m(vpeak)^2 so change in KE is KE final- KE initial } As you have used rms or average velocity ( you don't seem to be sure which it is) the numerical value that you get is too small by a factor of (2 or pi/2)^2 respectively). Again you have plugged into an equation without thinking of the meaning behind the terms. Note that in the next half cycle the change in KE is (0.5M(0)^2 -0.5M(vpeak)^2) which is negative It's gone- back to the system. Average energy in (and power) over a half cycle is 0 as velocity is 0 at the beginning and end. If you had attempted to sketch f and v vs time as suggested then found the power vs time curve, you would have seen this.

I only ask you to think critically. Is that too much?.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...

says...

instantaieous

frequencies.

+j(wM-K/w)and
Reply to
Don Kelly

Can't drop the insulting, can you Kelly? Grade-school behavior. Too bad.

This is absolutely disingenuous and dishonest.

You say my equation gives Pmec at resonance of: Pmec =11.12(0.157)^2 =0.274

I gave no such magnitude. My magnitude for net mechanical power at resonance was given as:

Pmec = Rmec fc * V^2 = 2.23 * (0.157)^2 = 0.055 watt

Now I suppose in your ignorance, you will apply this equation at 227.4 Hz and say "See I told you", but note again this equation doesn't work at 227.4 Hz. See below.

I understand *very well* how you see that as a problem. It took me a while to work out the relationship. I have given you the required procedure, but you ignored it, while hooting about my intellect, or some such crap.

You are disingenuous, Kelly. I told you velocity was average velocity. I've used the term v with a bar over the v. I've told you I calculated velocity as distance traveled divided by time taken.

Thank you for pointing out my error. The magnitudes of 0.027 and 0.0279 were close, and I became hasty. Since force is not in phase with velocity at 227.4 Hz cos angle must be used. Then for time during 1/4 cycle

Pmec = (1/2*m*v final^2*cos angle) + (v^2*Rms*t) / t =(1/2*0.0253*(0.0774)^2*0.313)+(0.0493^2*2.23*0.0011)/0.0011 = 0.0269 watt

Matching 0.0270 watt.

Note the second term on the right is the work done on the suspension and air load, with the first term being work done on the mass.

This should give rise to another long tirade on work done on the mass, since you can't understand it requires work to accelerate the mass above resonance. Considering this, I would not have replied in the last part above, except to correct my error.

Bill W.

Reply to
Bill W.

"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...

----------------- Sorry, my comments on non-answers, and illogical analysis, etc are valid. This is not an attempt to insult you. It is a statement of fact.

----------

------ I said that your "normal" method , applied at resonance, gave 0.274 watts

--------------------

---------- And I am glad that you have reduced this to what it should be rather than dealing with the pf and 1-pf nonsense. However, I say that Pmec =RmecV^2 is correct at all freqiencies where Rmec =2.23 no different calculation than at than at resonance..

-----------------

---------- Not ignorance - just using common sense and recognising where each term comes from.

----------

----------- Could it also be that you have made an error in the assumption that [(Bl)^2/Re]V^2 represents an actual mechanical loss? The problem is that such an instantaneous jump in power makes no physical sense at all. You have gone through a cut and try procedure using a "pf' and a 1-pf factor which you have finally realised is the application of a voltage divider and finally used Pmec (resonance) =RmsV^2.

----------------- In calculating velocity you used E=1.41 anf from this F=1.75 N which, among other quantities, you assured me were rms quantities. You have used an Fmax which agrees with this interpretation. However, the calculated velocity from the use of an rms force will be an rms velocity (0.0493m/s from the often repeated calculations).. You have also used P=RV^2 which is valid for the calculation of average power if V is rms. It is NOT correct if V is average. In other words, you have treated the velocity as rms, not average. I know that you indicated that the value of 0.0493m/s was average. I queried this and brought up these discrepancies with no response. You can't have it both ways -who's being disingeneous? Which is it - If it is average then a whole raft of calculations (yours and mine) and values such as Bl=7.17 are in doubt.

------------

----------- You are now getting more and more convoluted. You have made one correction but then apply a bugger factor with no reason except that it gives your desired results. The use of power factor here is completely invalid. I note that the pf you use applies to the whole of Fv in rms quantities, not the fv reationship for the mass alone (for which it is would be 0.) to get what you have predetermined is what you want. Did you ever think that this might be a coincidence? Try it with a different Re to see if the relationship holds-Say half Re. . Z=(7.17)^2/2.89 +2.23 +j33.75 =17.78+2.23 +j33.75 =20.02 +j33.75 =39.24 @59.32 degrees {magnitude 39.24] V=3.5/39.24 =0.0892 m/s rms but we will assume that this is average for calculating Vpeak to coincide with your calculation I will simlpy scale V=(0.0892/0.0493)*0.0774 =0.1400 m/s peak Now using Req=17.78 ohms and your approach to calculating Pmec =17.78(0.0892)^2=0.1414W Using your KE with power factor approach Pmec 0.5(0.0253)(0.1400)^2(cos59.32)/0.0011 +2.23(0.0892)^2 =0.225+0.0165 =0.2415 watts. Now I see no relationship between 0.2415 and 0.1414 watts. Try it yourself for another Re or with the same Re but at a frequency of , say,fc+0.1Hz, 60 Hz, 100Hz, 30Hz,1000Hz.- several different values. Have you tested your premise this way? Since these are the two approaches you used with only Re changed (by changing the size and or material of the winding but keeping the same length of winding, mass etc.) the relationship you claim doesn't appear. Yes, changing Re will change the Q and the half power points so the response will change to a sharper resonant peak. This doesn't support your contention. The power factor you have applied is dependent on the phase angle of the total impedance including the M, (Bl)^2/Re ,Rms, K and w. Plugging it into the KE expression isn't valid. Please note that the change in KE depends only on the final and initial velocity magnitudes. In the quarter cycle, the initial velocity is 0 and the final is 0.774 m/s. The change in KE is 1/2M(vfinal^2 -Vinitial^2) and there is no angle or average involved with these velocities which are the end point values which in this case are the instantaneous velocity magnitudes at the start and end times. No phase involved- No power factor. Hence in this 1/4 cycle, the change in KE is 1/2(0.0253)(0.774^2 -0) =7.6x10^-5 Joules. The average power into the mass is then 7.6x10^-5 /(0.0011)=0.069 watts. Now go to the next quarter cycle: initial velocity =0.0774, final velocity = 0 change in KE =1/2(0.0253)(0-0.0774^2) = -7.6x10^5 Joules That's right - negative. Energy is being returned to the system. power into mass = -0.069 watts Reversal of energy change implies a reversal in power . What went in during the first 1/4 cycle, comes out in the next 1/4 cycle.

Alternatively go over a half cycle Vinitial =V final =0 Pinto mass = 1/2(0.0253)(0-0) =0 I notice that you chose not to address this and even snipped it without noting that you did. . This is the crux of the matter, not some relatively minor numerical error and invalid corrections. No matter how you cut it, if you honestly do the analysis, the result is the same. That is why all the literature treats mass as a reactive element rather than resistive. The force on the mass is accounted for in the use of F=ZV and the component of F applied to the mass is reactive - that is the force acting on the mass is 90 degrees out of phase with the velocity. (cos

90 =0)

------ Agreed as to the second term but not as to the first term.

------------ You only "corrected" one error which was realtively minor. Your basic premises are the major errors and you have not addressed these premises or criticism of these premises. They are: a)The electrical term Re and the flux density B as well as the conductor length somehow produce an real rather than equivalent mechanical impedance and losses in the (Bl)^2/Re are actual mechanical losses. You have ignored the development of the equations which quite simply give rise to this term. Instead you have repeated equations with your numbers and claiming that is what your sources do. One at least certainly doesn't (and I suspect that on rereading , the others also do not support this) and circuit theory as well as common sense do not support this. b) In order to try and explain this, you have come up with a power to accelerate the mass. Again the explanations are rather feeble and full of inconsistencies. You base all on one 1/4 cycle without examining what is going on in the next 1/4 cycle. Even when challenged to do so, you have made no attempt. . If you have a good theory then you should try to destroy it - i.e. test it critically - rather than simply ignore criticism. That you haven't done. You haven't made any attempt to understand the criticism or the rationale for this criticism. - at least make a reasoned argument. Repeatred statements which claim that I have said something which I haven't doesn't make it true except in politics. c)You make a big point of the difference between at, and above resonance - recognising a discrepancy and I give you credit for that but then finding about as awkward a way as possible to explain it away. d) Discrepancies crop up everywhere but again these are not addressed- such discrepancies are signs that all is not well and possibly, just possibly, a basic premise is faulty. Simply ignoring discrepancies doesn't make them go away. Others are related to what terms are average and what are rms as pointed out above. Apologies for a long tirade.- would you rather that I make a very terse (say one word) summary? Would that help?

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

Then you are either too ignorant to recognize an insult, or insult purposely, then claim you didn't, making you either ignorant or untruthful. Only you know which.

This statement itself makes you disingenuous. I gave no "normal" method showing Pmec at resonance =11.12(0.157)^2 =0.274. Give reference (Google works well) where I did such, otherwise you are disingenuous at best. It will be interesting to see you wiggle around this. I'll be waiting. Clip and wigggle about now?

1>> See below.

Beranek defines (Bl)^2/Re as a mechanical resistance. Do you question this? Yes or no?

Mechanical resistance times velocity squared equals dissipated power. Do you question this? Yes or no?

Beranek also states (equation 7.2) Rmec above resonance = (Bl)^2/RE + Rms + Rmr assuming negligible source impedance. Do you question this? Yes or no?

There is no instantaneous jump in power, and I never stated there was. My equations were either for power at resonance *or* for power at 227.4 Hz. The best I have at present is that the transition takes place either proportional to or as the square of back emf. Note that BLv at 227.7 Hz / BLv fc = 0.313 = PF.

-----------------

Look Kelly, enough is enough, OK? Look at what you replied to above. Here, read it again............

"I told you velocity was average velocity. I've used the term v with a bar over the v. I've told you I calculated velocity as distance traveled divided by time taken. "

Now the fact that other calculations match my measured magnitude is skin off your butt, not mine.............

Really, stop the disingenuous crap, OK?

What is in doubt is your magnitude for mechanical power of 0.0054 watt at 227.4 Hz

I have addressed your major error. As I have noted, no point in so many details until it is cleared up.

. This is the crux of the matter, not some relatively

You do not realize that the mass is mainly the load, nor do you realize it requires energy to be accelerated above resonance. You just hoot about what adds in one time frame cancelling out in another, but............. *****COMPLETE***** cancellation of mass reactance occurs **********ONLY AT RESONANCE**************** I suggest you consider this seriously, otherwise we are wasting out time. As you say "Is that too much to ask?"

I suspect you refer to Kinsler as your support. I requested the page number and you spewed out another long dissertation. Do this please:

Give me the source (Kinsler or whoever) and the *equation number* that allows mechanical power above resonance of your magnitude of 0.054 watt. I have provided many that give my magnitude of 0.027 watt.

Until you show support, as I have, you are looking much like the proverbial dunce here. And I know you don't want that, so just give the *equation number* that supports your 0.0054 watt.

Ahhahahaaha sorry.... really that's funny.... explain that we need power to accelerate the mass, really,, you're not kidding are you hahahahaaha again soerry . .

You base all on one 1/4 cycle without examining what is

It's you projecting, not me Pal.

I hate one-word conversations. Could you just find a middle of the road? :)

Bill W.

Reply to
Bill W.

-----------miscellaneous nonsense on both sides snipped.---------->

-------------- I note that in your "correction" you did not challenge this, being upset about a digit transposition. To be fair you have only applied this approach at frequencies not equal to resonance and appplied the correct method at resonance.

----------- I don't know what Beranek said. I only know what you say he said. I do agree that this term is an "equivalent" mechanical resistance representing the effect of Re transposed to the mechanical side. I have no problem with this and have said so repeatedly. However, you don't say whether Berenak uses this in calculating the "actual" mechanical power. I would like to know what he actually does, not waht you do.

-----------

--------- No. -IF the velocity is rms. otherwise a bugger factor must be used.

--------

----------- No problem. Please note that the "above resonance" restriction is not needed.

-------------

------------- I should have said "discontinuity" rather than jump. Have you attempted to calculate the velocity and power at other frequencies such as at 100Hz, 60Hz, half power frequencies, (1.01fc), (1.001fc), (0.99fc), etc? I suggest that you try it. Please note that as you approach fc the velocity approaches 0.157 m/s and the power factor approaches 1. If your equation for power "above resonance" is correct, then as you sneak up on resonance. the "power " approaches

11.12*(0.157^2). However at resonance it is (correctly) 2.23(0.157^2) =0.055 watts. This is a discontinuity. There shouldn't be such and will not be such if you recognise that the approach you use at resonance is correct at all frequencies. E-ReI =Eb leads to I=(E-Eb) /Re or F=BlI =BlE/Re -BlEb/Re =BlE/Re -V (Bl)^2/Re Note that this is simply an expression for the Lorentz force. As such, it doesn't involve then mechanical parameters at all, nor is it specific to any given frequency - It is simply a Norton equivalent source as seen from the mechanical side. No more than that. I really don't know what all the fuss is about. The mechanical system gives F=BlI =ZmV (Zm doesn't include (Bl)^2/Re) and equating these leads to a solution for V. It is convenient to think of (Bl)^2/Re as a mechanical impedance for this purpose but not for the purpose of calculating mechanical power.

As for BlV at 227..7 Hz/BlV fc =0.313 I should hope so - implies that resistance is constant. BlV f/BlV fc =V f/V fc =Zfc/Zf =11.12/root (11.12^2 +X^2) = the pf at frequency f. So what's the point?

----------

-------------- You are avoiding an issue. I only note that *IF* E=1.41 is an rms voltage, then F=1.75 N is an rms value and F/Z gives an rms velocity. We have both used these values, treating V as rms and the calculations are consistent. If E is an average value, then I have no problem with V being an average value, but you assured me, way back when that E (and I ) were rms as would be expected. Either these are all rms or all average, one or the other- not a mixed bag (which would require two different Bl terms) . I queried this before- no response. . In power calcuations you treat V as if it were an rms value, not an average value. Using P=R(V^2) is correct for an rms V but is INCORRECT for V average (except for DC or a square wave where rms=average) - for a sinusoid, this leads to a 23% error.

------ >

------------- Oh, no, there is no doubt. I have interpreted it correctly.

----------

--------------- Yes -the major error was a numerical transposition. Corrected, thank you. However. the part above that you so studiously ignore is not a mere detail but the whole monty. If what I have said above regarding KE is correct, then your acceleration power concept is blown to ratshit. If it is in error, then I expect you to analyse it and point out the errors. (I am deliberately avoiding the use of calculus as it doesn't appear to be in your tool box) I suggest that you read it and present a counter- argument rather than avoid the issue. As for testing your concept at different frequencies- why not do it- in spite of my suggesting it- see for yourself.

---------

---------- I have considered it seriously. You haven't - note "mass reactance" not "mass resistance" - there is a reason for that. The mass does affect the force/velocity relationship but it only involves the out of phase component. The mass and spring compliance reduce the power factor. Neither increases the real power . Also note that the EbI cos (angle between Eb and I) leads to an "airgap" power (power converted from electrical to mechanical power) is less than the value (11.21)V^2 so your approach leads back to a convwersion efficiency greater than 1. This is a non-trivial point. >

------------- Kinsler pp347,348, 351-3,360-61, and following Fig 14.3,14.5, 14.10 Eq.14.9-11, 14.23-30, 14.48,14.51 etc. Note that I gave you a detailed calculation using Kinsler's approach with Zmot (fig 14.10 and eq14.51b) at both 227.7Hz and at resonance and the results agree completely with the results that I presented earlier.

----------

---------- I have invited you to examine my analysis and criticise it. It's up to you. Hahahaha is a polite form of Darkmatter's response to questioning of concepts. You don't want to be in his league. If you are willing to do this- fine, otherwise, no skin off my nose. As I said before, all I ask of you is that you think. You have a concept- I have criticised it- really quite gentle criticism compared to what yo would get in a conference. -- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

------------

resonance -

Reply to
Don Kelly

So you cannot back up your false statement with a reference. Too bad.

A non-answer.

Then if you have no problem with "Rmec above resonance = (Bl)^2/RE + Rms + Rmr" you should use the term for Rmec at 227.4 Hz and correct your erroneous power equation.

I was hoping to steer you toward solving our difference.

No, you are the one avoiding.

I only note that *IF* E=1.41 is an rms voltage,

How many times do I have to tell you, I will not address your sidestepping such as the above, until you correct your power equation. No point in you hooting about ~ 10% error due to RMS vs average, 23% error as the above, etc etc when yout power equation is off by about 500%.

No, you have not, you have offered no support. See below.

That's getting to the heart of why I came here. Consider this carefully, please:

Mass is a measure if inertia, that is, of the *resistance* to a change in motion.

If you cannot see there is a resistance (real) from the mass do being accelerated, and further that the power required to accelerate the mass in a given quadrant of a cycle *above* resonance is more than the power returned in the following quadrant of a cycle then we may as well button it up.

Another red herring. They have become tiring.

NONE of these are power equations.

ONE MORE TIME, GIVE A POWER EQUATION THAT SUPPORTS YOUR Pmec = v^2*Rms = 0.0054 watt

Kinsler gives them up front in section 1.9. You have the book. Use it. Spell it as I have for mine, i.e.

Per Halliday: Pmec = F*v*cos angle = 1.75*0.0493*0.313 = 0.027 watt.

If you cannot do this, then clearly you are wrong, and your analysis is wrong.

No, it's up to you to offer support for your power equation, per above. Failing that you lose the argument, and I'm sorry it had to come to this. I really expected more here - had hoped for much more. A major dissapointment. Not much more I can say.

Bill W.

Reply to
Bill W.

--------- I went to a copy of previous correspondence and did quote back your "correction" I did not bother going to Google. Nor will I. What I have said above is correct.

-------------------

--------------- OK - a qualified Yes. The qualification is "equivalent" We have gone over this before. Have you stopped beating your wife? Yes or No?

------

------------------ I use this term and have used it right from the beginning. when I used V=F/Z =1.75/(11.12 +35.54)=0.0492 @ -71.76 degrees. I have told you this again and again. I also have told you again and again, that I do not use it for calculating the mechanical power and also the reasons why I don't use it. How many times do I have to do it?

You still don't seem to get it.

Here's another thing- If Berenak actually uses [(Bl)^2/Re + Rms+Rmr] to calculate the mechanical power, rather than just the velocity, (and I note that you didn't answer this) then he is *wrong*. Clear enough? I doubt very much that he is stupid enough to make such a fundamental error.

------------

----------- Then look at what I said and critique it point by point- not by quotes but by thinking. Look at the equations and look at what each term means and where it comes from. So far you have said Berenak says ... with your interpretation or Halliday says ... and giving a very basic equation with your numbers (which he didn't say) thrown in. Do they explicitly calculate a power using the (Bl)^2/Re term or BlE/Re as the force.? I agree with the use of these to determine V but not for power calculations- not because someone said so but because I can see where these terms come from and what their physical meaning is.

------------

---------------- This is not sidestepping-either you can answer or you cannot. I really don't give a damn at this stage.

Compare your power to the airgap power determined from EbI cos(angle between Eb and I). Also find the input power less the I^2Re loss ( something I did and got the same airgap or total mechanical power that I got from RmsV^2 ) At least my results do satisfy conservation of energy. Have you checked yours for this? What Eb do you get, what I.? what relative phase. Again, Your results are wrong and it likely doesn't matter because you probably use a correct formula to get the efficiency (or lucked out in the use of the formula) and actual acoustic power and the mechanical power is really not of concern.

In other words, I am saying that the resistance you use in calculating power is wrong. The average vs rms etc is simply mathematical sloppiness on your part. The not realising the differences as appears to be the case is worse than sloppiness. ----------

------------------ Actually, I have. I havent done it by quoting specific equations, etc but I started with the basic electrical and mechanical equations and developed the model from there. All was laid out step by step. No other support needed. It is all there. I gave you credit for the ability to read and think. I did not jump in with " so and so says xxxx" without giving being clear about all the terms in use and why they are used as you have done. I gave you a road map - it has been obvious from the start that you have had problems with it. I suppose that, if I quoted someone at every trivial step, you would be happy. Forget it. If you find fault with the work I have shown, then criticise it - you picked up on a discrepancy with your conceptions and yelled about it but didn't bother to look at what i did and where the difference came from. The formula is minor- it is what leads up to it that is important - that you haven't shown- simply quoting someone is NOT support. It simply means that you can copy an equation and plug numbers into it - You have given several equations to "prove" something and I have been able to determine how they arise and what they mean. Later I looked at Kinsler and used his methods to get identical results to mine from a different approach.

--------------

=0.225+0.0165

------------- In physics and engineering "resistance" is used for dissipative elements Mass isn't one of these. You say that the power to "accelerate" is higher than the power returned but you have given absolutely no reasoning for that. Show your reasoning and the physics behind it. Handwaving as above doesn't count. You use KE first quadrant : change in KE =(1/2M(Vmax)^2 -0) =+7.6x10^-5 Joules second quadrant change in KE =(0 -1/2M(Vmax)^2 )=-7.6x10^-5 Joules sum =0 Power =0 Over the half period initial and final velocity =0 so change in KE over half period is 0. Power =0 Typical of any reactive element - no dissipation. If this is wrong - point out where and why.

-----------------

-------- Not a red herring - The "air gap" power is the power transferred from the electrical to the mechanical system. this is fundamental to any electromechanical device. It is related to conservation of energy which is very definitely not a red herring.

--------

--------- That's right. So what.

-----------

------------------ The equations in Chapter 14 aren't power equations- true. However, somewhere along the line he has assumed that the reader understands the concepts of power, instantaneous and average. I should have thought that the equations were obvious and, in fact, I used the same equations that you keep throwing out as gospel but with a different force or resistance. Quoting an equation won't change this. You still act as if Halliday etc are using your numbers and your interpretation. I am happy with their equations as I told you before- they work for me.

There are many ways to express power. Note that Kinsler's power equation (based on maximum F ) P=F^2/2Zm cos theta is also = Frms^2/Zm cos theta =FrmsVrms* where Frms and Vrms are phasors and Vrms* is the conjugate of Vrms. He doesn't repeat these equations in chapter 14- there is no need to do so. I note that Kinsler uses Zm and also note the paragraph below which leads to (F^2/2Zm^2}Rm which becomes V^2Rm. Take note of this paragraph which notes that the "average power supplied to the system is not permanently stored in the system but is dissipated in the work expended in moving the system against the frictional force Rm u" This supports my contention but cerainly doesn't support your acceleration power contention. Further note that in chapter 14, his use of Rm agrees with mine (he uses the "open circuit impedance" which is the same thing) and does not include Bl^2/Re as he takes Re into account in another way. His circuit model can easily be shifted to the mechanical side where it becomes the one that I used. .

----------

---------- As I told you before, You can plug in a set of numbers into a formula - the formula doesn't care so if you plug in garbage, you get garbage. That is what you have done. Quote and plug without thinking of the basis of the expression:

Per Halliday: Pmec =F*V*cos angle1.664*0.0492*cos(86.22)=0.0054 watts This can be written as Pmec =V^2Rms =(0.0492^2)2.23 =0.0054 (you use the same equation with an R of 11.12 while I use the actual mechanical resistance ) Same equations that you use but the force is the actual force , not the locked cone force which doesn't exist except at v=0. and the impedance is the actual mechanical impedance omitting the term depending on the electrical resistance Re. What's the point of quoting formulae? Understanding them- that is the rub.

In addition- some other power expressions which are quite valid

EI cos(angle) -I^2Re =EbI cos(angle between Eb and I) =0.232*0.353*cos(-76.76-14.46) =0.0054 watts

Have you tried the latter calculation?- it is the same as Halliday's except in electrical quantities. I have done this before- at least twice, and have also used Kinsler's approach (which doesn't include (Bl)^2/Re) as all is referred to the electrical side and obtained the same results. Somehow I have consistent results. Do you? You seem to want me to quote you a power equation. Sorry, but equations for power are ingrained from use. I don't need to get a specific reference any more than you need to give a reference for Ohm's law. I find nothing wrong with the ones you have given or the ones in Kinsler. just plug in meaningful data to get meaningful results. Try it. There are also some In Gourishankar & Kelly, "Electromechanical Energy Conversion", White and Woodson, "Electomechanical Energy conversion", Krause "Analysis of Electric Machinery", plus those in any circuit texts you want to use.

------------

----------------- Cute but asinine. The problem is not in an equation for power but in the understanding of what terms are used in the equation. You have again avoided the issue.

Remember this: I= (E-Eb)/Re (electrical equation) or BlI =(BlE-BlEb)/Re which becomes using F=BlI and Eb =BlV F=BlE/Re -[(Bl)^2/Re]V =1.75-8.89V = BlI**** This has nothing to do with the mechanical impedance but is simply the electrical equation expressed in mechanical terms. (Bl)^2/Re is an admittance which looks like a mechanical resistance - no more than that. Do you have a problem with the above? Yes or No? If so where and why. It goes to the heart of the problem. As yet, you haven't, in spite of quoting Halliday, etc, (outof context) given any indication that they use your interpretation.

If you bother to get by that, then you can look below

We can also see that EbI =FV so another power equation is P=EbIcos (angle between Eb and I) or, since Eb =E-RI this becomes P=EIcos (angle between E and I)-I^2Re The results of the use of these power expressions should agree with P=RV^2. If not there is a serious problem.

The mechanical equation is F =[Rms +j(wm-K/w)]V =ZmV where Rms is the actual mechanical resistance so 1.75-8.89V = BlI = (2.23+j33.75)V or 1.75 =(11.12+j33.75)V is convenient for finding V but that's all it is good for.

As a last comment: I note that you don't ignore the (Bl)^2/Re term at resonance. It is there Vfc =1.75/11.12 =0.157 m/s

This is correct. It is there and you use it in calculating power at all other frequencies. You don't do this but you then, correctly, ignore it in calculating power. To be consistent, should you not use Vfc =1.75/2.23? or use P=2.23(V^2) at all frequencies? There is no reason for a special relationship at resonance as there is nothing really different in the governing equations at resonance except the magnitude/sign of the reactive term (going from - to + ). It is an important point but different to the extent that it requires a special correction- hardly.

Surely you can do better than the above.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

I ask you for Kinsler's power equation that supports your Pmec = v^2*Rms = 0.0054 watt at 227.4 Hz. I ask you to spell it out with magnitudes as I have, i.e. Per Halliday: Pmec = F*v*cos angle = 1.75*0.0493*0.313 = 0.027 watt. You gave non-power equations and I complained noting they were not power equations. You replied:

Enough said. You have the book, and I even told you the section Kinsler's power equations are in. However... you do not show a power equation with magnitudes to match your mechanical power of 0.0054 watt at 227.4 Hz, because there is no such equation, so you lose the argument. Now in the following you attempt to manipulate Kinsler's actual power equation and continue your evasionary tactics to avoid admitting your error, however your rambling and blame shifting doesn't mean crap.

------------------------

Concerning power at resonance, I said: "So you cannot back up your false statement with a reference. Too bad."

You replied: "What I have said above is correct."

Therefore you were either untruthful or you misunderstood and mis-stated my equation for power at resonance and you lack the courtesy to go back and correct things. Neither speaks well for you.

-------------

-------

So you offered your own support. That would be funny, were it not so pathetic.

Yes, but your different approach gives a different power at 227.7 Hz than his power equation - off by over 500%. You know this, but won't admit your error. Again, this doesn't speak well for you.

= (0 -1/2M(Vmax)^2 )-7.6x10^-5 Joules sum =0 Power =0

You are correct for power accelerating mass at **resonance**. I was incorrect re quadrant power. Sorry.

One of the most blantant cop-outs I've ever seen.

Asinine, eh? Candidness doesn't mean crap to you compared to your self-aggrandization, does it?

That you are desperate, is shown by your following diatribe in your ongoing attempt to evade acknowledging your error. Bill W.

-----------------------------------------------------

-----------------------------------------------------

Reply to
Bill W.

--------------- Oh, my, How about looking in Kinsler- in the page range that I originally quoted. eg. P361, text below Eq.14.53 "...the total power consumed is RoI^2 +(Rm+Rr)u^2 ..." In the terms we are using this becomes Ptotal =ReI^2 +RmsV^2 The first term is the electrical loss and the second is the loss in the mechanical resistance + acoustic output which is exactly what I have been saying all along.

Note also that this power relationship holds at all frequencies. It is implicit in the efficeincy expressions but is implicit in the whole development of Ch.14. Resonance is simply a specific case which while important, does not require any difference in the approach to calculation of velocity, current, power or whatever.

The efficiency that he presents in Eq 14.54a can be found from RrV^2/Ptotal ( he just re-expresses I in order to eliminate V) This reduces to the expression in 14.54b at resonance but is no more than the same equation in another form.

Note also from the definition of Zmo (in eq 14.1) there is no term related to Re as with I=0, there is no electrical coupling - as if the coil wasn't present. The F in the canonical equations (14.2) is an applied mechanical force (mice?) Also note the equations 14.22 and the descriptive text following. All this is in agreement with what I have been saying.

Three factors are apparent: a) power goes to I^R loss , mechanical losses depending on the mechanical resistance and acoustic output - both of which are form RmsV^2 No power to accelerate the mass exists. b)No difference occurs at, below or above resonance. The same equations hold. c)I didn't twist Kinsler's power equation- This and the other forms that I gave are something that I have known and used for 50 years and find it quite unnecessary to quote . After all they do come directly from p(t)=f(t)*v(t) at any instant of time t. (and Kinsler uses this form for f(t)=Fcos wt and v(t) =|F/Zm|cos wt-theta). The last paragraph of section

1.8 is also of interest. I also suggest that you look at section 1.7 which may help with regard to the response of the mechanical system.

----------------

acceleration

------------- The equation for power at resonance is the same as at any other frequency. There is no special consideration needed. This is all going back around the loop again. Read again what I sad- don't twist it. Your power at resonance is correct- I also said that. I also said that the power calculation at any other frequency is the same (pmec =RmsV^2 ) (see above)

------------ I'm sorry that you couldn't follow it - you apparently think it is better to quote blindly than to develop from scratch. Too bad. I prefer to see where and how the relationships are developed rather than simply take them on trust. I presented the steps - you are free and welcome to go through them and look for errors in analysis. I don't want you to take them on trust. However, If your counter argument is weak then I will counter it - and that has been the case so far.

-------------

------------------- Excuse me? Where do you get this? You obviously only read what you want to read. I got the same power and other factors using his approach that I got using mine. You have what I did both ways Zm=2.32+j33.75=33.82 @86.22 Kinsler: Zmot =(bl)^2/Zm =[(7.17)^2/33.82] @ -86.22 =0.10 -j1.517 Ze =5.78+Zmot =6.07 @ -14.46 degrees I=1.41/Ze =0.232 @ 14.46 degrees You saw this before as you questioned the 0.1 ohms and indicated that your Ze was 6.21 ohms etc, (which side of the ammeter did you measure voltage?) Now the total resistance is 5.88 ohms of which 0.1 ohms is reflected from the mechanical side. The power in this is 0.1(0.232)^2 = 0.0054 watts How is this different from 2.32*0.0493=0.0054 watts?

Alternatively Pin =EIcos angle =1.41*0,232*cos 14.46 =0.3168 watts I^2Re =5.78*(0.232)^2 =0.3111 watts Pin-I^2Re =Pmech =0.317-0.311=0.0056 watts (loss of accuracy from roundoff)

----------

------------ Note that the KE argument is valid at all frequencies, not just at resonance. Do you see anything in the expression for change in KE that depends on resonance? It depends only on the mass and the initial and final velocities.

-------------------------

---------------- No- it's just that simple.

-----------

----------------- Now who's getting uptight? How is it that whenever I ask you to explore something for yourself or to actually respond with intelligent criticism of what I have said (as below), you accuse me of wriggling. If what I have said below is in error , point out the errors. No desperation, no lack of candour- I just gave some relationships and would welcome well thought out criticism (the above is certainly not ). I have asked you this before. I have laid out the steps that I used so that you could go through them and question them. I have also pointed out discrepancies that have arisen in what you have presented and waited in vain for a response to these questions. No big deal, no glory. Just wanting you to think, not plug and pray. At least try to come up with good arguments.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

Oh my, indeed... You refuse to use the actual power equations again, plus by not being handy with figures, you shot yourself in the foot with your above equation. I love it... grade-school level, was it you said? OK, you quote above: power = RoI^2 +(Rm+Rr)u^2 Ro = Re = 5.78 I = 0.227 Rm = 2.23 (YOUR mec resistance at 227.4 Hz) Rr = 0.629 u = v = 0.0493 5.58*0.227^2 + (2.23+0.629)*0.0493^2 = 0.305 watt

Off from IE = 0.227*1.41 = 0.320 by 4.7%

Now with my plan using (Bl)^2/Re = 7.17^2/5.78 = 8.89 for Rm so as to fit into your handy-dandy equation...

5.58*0.227^2 + (8.89+0.629)*0.0493^2 = 0.321 watt

Off from IE = 0.227*1.41 = 0.320 by only 0.31%

Your error being 15 times that of mine in percentage.

This shoots your equation all to hell. So you still lose the argument. But....... you can still do the honorable thing by admitting your error and quoting Kinsler's POWER EQUATION (page 14), along with magnitudes. That would still lose the argument for you, but it might get you back in my good graces...

A question: Why do you NOT use Kinsers power equations on page 14? The answer should be *very* interesting. Bill W.

---------------------------------------------------

---------------------------------------------------

Note also that this power relationship holds at all frequencies. It is

Reply to
Bill W.

-------------------- What a pile of crap. First of all you now add the term 0.629 for Rr which was , as I understood, rightly or wrongly was supposedly part of Rms as you gave it. I was using only the Rms that you gave me- that is 2.23A. You have not used the 0.629 before now. (remember 8.89+2.23 =11.12 has been used until now in your calculations and in mine.) That in itself is not a problem. The two problems are below

1) Now you are using 8.89 +0.629 as the Rm +Rr. How droll. Why not 11.12+0.629-what happened to Rms? Note that the 0.629 will affect the velocity - so it will not be the 0.493 m/s that we both found from (BlE/Re)/Z =1.75/(35.54) =0.0493 m/s but something lower. 2) You have also assumed E and I are in phase- not so- you are sliding back to a bad habit.

Suppose we go back to the original numbers assuming Rr is included in the

2.23 as we both have done before- mainly because I am not going to recalculate in detail at this time. I calculated I=0.232 @14.46 degrees. I know that this is different from what you have but it is found using data that you provided. Then P=0.232^2 (5.78) + 2.23(0.0493)^2 =0.3111+0.0054 =0.3165 watts Now, if you were consistent you would use 8.89+2.32 = 11.12 as your Rm (as you have done all along) and the result, using my current would be 0.232^2(5.78). +11.21(0.0493)^2=0.3111+ 0.027(remember this?) =0.338 watts 2) EI is not power as E and I are not in phase. Taking into account the phase I get P=EI cos 14.46 =0.3168 watts Which agrees quite well with what I had.

Using I =0.227 magnitude (and the phase 21.45) won't help either This has all been done before. except for your addition of terms previously ignored, ignoring of phase information or omission of values previously used.

Adding the 0.629 and correcting the velocity accordingly will result in a velocity of 0.490 @-70.8, current of 0.231 @ 14.35 so the powers above will not be appreciably different (Pin=0.3155: I^2=0.308+, ms+Rr)V^2=0.0069 -checks out. )

------------

------------------ As a matter of fact, I have. You simply seem to have a problem in expressing them in different forms.

First of all, I pointed out that the basis of this equation is simply p(t)=f(t)*v(t) (read the first paragraph of section 1.9) expressed in terms of Fpeak for a sinusoid and an impedance- hence the F^2/Zm cos(wt)*cos(wt-theta) {theta is the phase angle associated with Zm} He then integrates and averages to get Eq1.34

Note that V=F/Zm gives FV =F^2/Zm as used. Noting that Kinsler uses peak F (=Frms*root(2)) in this equation Eq1.34 with rms values becomes Pave=(Frms^2/Zm)*cos theta Eq 1.35 becomes Pave =Rm *(Frms/Zm)^2 Since V =F/Zm , F^2/Zm =FV (F/Zm)^2=V^2 so Kinsler's equation 1.35 now can be written as Pave =Rm*V^2 which is what I used (valid for rms V -divide by 2 if peak V used)

It also becomes P=FVcos theta as per your other references. (valid for rms F and V- divide by 2 if peak F,V used) These are all ways to say the same thing. I use the form which I find most convenient at the time. No big deal.

However, here you are from a plug in and turn the crank approach: Pave =(Frms^2)Rm/Zm^2 =1.664^2 *2.23/33.75^2=0.0054 watts

I have used the actual force =BlI , not the force at V=0 (for which the mechanical power will be 0!)

Kinsler's first chapter 1.1 to 1.13 is a review of freshman to sophomore level material ncluding a bit on phasors and circuit equivalents.

Sorry, you will have to do better than this.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

I hesitate to call someone a liar, but you have no regard for accurate or honest communication. You state below: "You have not used the 0.629 before now." It was used as part of Rms and you were told this on 1-20-2004 in an explicit manner.

-------------- reference --------------------

formatting link
Or search Google for "air load resistance as was measured", where I stated:

"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"

------------ end of reference -----------------

You ask below " Now you are using 8.89 +0.629 as the Rm +Rr. How droll. Why not 11.12+0.629-what happened to Rms?

I'll spell it out for you. You quoted: power = RoI^2 +(Rm+Rr)u^2 Ro = Re = 5.78 I = 0.227 Rm = 2.23 (YOUR mec resistance at 227.4 Hz) Rr = 0.629 u = v = 0.0493 5.78*0.227^2 + (2.23+0.629)*0.0493^2 = 0.305 watt Off from IE = 0.227*1.41 = 0.320 by 4.7% Using my magnitudes where Rmec = (Bl)^2/Re + Rms = 8.89 + 2.23 = 11.12 (note Rms includes Rmr = 0.629) 5.78*0.227^2 + 11.12 *0.0493^2 = 0.325 watt Off from IE = 0.227*1.41 = 0.320 by only 1.56% You miss the mark by over 3 times that I do.

Reply to
Bill W.

------------- If you cannot follow the manipulations- that is not my problem. Kinsler uses peak F in his equation. I pointed out that Frms for a sine wave is Fmax*root(2) {or Frms^2 =Fmax^2/2}so that the power P=(Fmax)^2 Rm/2Zm^2 => Frms^2 Rm/Zm^2 Do you have a problem with that? It still comes out to 0.0054 watts. See below

----------------

formatting link

G=Google+Search&as_epq=air+load+resistance+as+was+measured&as_oq=&as_eq=&as_ ug

=&as_drrb=q&as_qdr=&as_mind=12&as_minm=5&as_miny=1981&as_maxd=7&as_maxm=2&as _m

----------------- This is what I thought. As I said as you can see below. "First of all you now add the term 0.629 for Rr which was , as I understood, rightly or wrongly was supposedly part of Rms as you gave it. "

You have not *used* it as a separate entity before (and I expected your response) . You have included it in the Rms =2.23 ohms. My understanding was correct as you have confirmed. Therefore why are you now using it as a separate element if it is included in Rms and ignoring the rest of Rms ?

******** and Kinsler's - at any frequency assuming Rms is a constant. *******
*** But you say that Rms includes Rr so whay are you adding it? ****
****** I and E are not in phase so IE is not power. *********
***** You have made a correction -good. This is not what you said originally. However you now have two terms which depend directly on Re. How come you can ignore change the Rmec to 2.23 from 11.12 at resonance (your previous arguments are not valid so try something else). In addition, the approach that Kinsler uses (and these calculations are based on Kinsler's model of Ch 14) the term (Bl)^2/Re doesn't appear at all. His Rm is the actual mechanical resistance. That is a point I made before. However, you still haven't corrected IE for the power factor. Do it. . I simply pointed out : 1) the use of the 0.629 term separately from Rms. Up to now you have used Rmec =11.12 which includes the Rms which in turn includes the Rr term. and, by your calculations, have claimed a mechanical power of 0.027 watts.You have conveniently changed this to 8.89 +0.629 and simply snipped my comments and correction. 2) the omission of the rest of Rms in what you have done. 3)the ignoring of the phase angle of the current. and presented corrected values. I quote , "Suppose we go back to the original numbers assuming Rr is included in the 2.23 as we both have done before- mainly because I am not going to recalculate in detail at this time. I calculated I=0.232 @14.46 degrees. I know that this is different from what you have but it is found using data that you provided. Then P=0.232^2 (5.78) + 2.23(0.0493)^2 =0.3111+0.0054 =0.3165 watts Now, if you were consistent you would use 8.89+2.32 = 11.12 as your Rm (as you have done all along) **** This is what you now do ******* and the result, using my current would be 0.232^2(5.78). +11.21(0.0493)^2=0.3111+ 0.027(remember this?) =0.338 watts You get 0.320 using a current of 0.227A Not bad but coincidence as the dominant term is I^2R - you have a lower I and a higher mechanical resistance-. compensating errors. 2) EI is not power as E and I are not in phase. Taking into account the phase I get P=EI cos 14.46 =0.3168 watts Which agrees quite well with what I had."
Reply to
Don Kelly

I can follow logical manipulations. The real problem is that *you* cannot or will not follow logical manipulations set down by authorities in the field concerning mechanical resistance and mechanical power.

An example of your abounding sloppiness. Rms = peak * 1/sqrt2.

I have no problem with Kinsler's equation, but I do with the mechanical power magnitude you use and so does Halliday and Serway. You give Kinsler's eq. 1.35

P=(Fmax)^2 Rm/2Zm^2

First, using *your* Rmec = 2.23, to calculate mechanical power at 227.4 Hz and with Rms force converted to average as 2.75 * 0.637/0.707 = 2.475 P = 2.475^2 * 2.23 / 2 * 35.53^2 = 0.0054 watt

Corresponding to your claim.

---- Now lets use Halliday's, Serway's, and my magnitude for Rmec of 11.12 to calculate mechanical power at 227.4 Hz, again with Rms force converted to average as 2.75 * 0.637/0.707 = 2.475

P = 2.475^2 * 11.12 / 2 * 35.53^2 = 0.027 watt

This matches my claim of 0.027 watt, not your claim of 0.0054 watt.

We've been here before - took you several days to come up with a weaseling response as I recall..

Lack of response to your Google-proven error speaks for itself with reference to you not acknowledging your mistakes. Too bad.

Example of your weaseling - you now add the word "separate", saying: "You have not *used* it as a separate entity before"

A taste of your own medicine?

A taste of your own medicine?

See below.

I have told you, an appropriate equation will give Rmec = 11.12 at 227.4 Hz and 2.23 at resonance. I even told you I was working on the transition rate empirically. This is the part of your lack of comprehension.

I believe I ask you to quote where he states this and you never responded. In any event you are dead wrong. Kinslers mechanical resistance equation IMMEDIATELY FOLLOWING his power equation 1.35 as discussed above is

F Rm = --------------- cos wo t = 11.13 v resonance

Which is NOT actual mechanical resistance of 2.23.

In the equation: F = 2.75 v res = v res avg*pi/2 = 0.157*1.57 = 0.247 wo = 2 pi fc = 2*3.14*58.6 = 368.2 t = 1/4 cycle at 58.6 Hz = T/4 = 0.00427

Note if F is factored by 0.9 then so must be your Rmec of 2.23 (reduced by 10%). Regardless, what is 10% compared to your error of 500%, i.e. 2.23 vs 11.12? Therefore I will not respond to a windy disertation regarding Rms vs average unless you admit your 500% error.

?Do it.

Power in either equals power out or it doesn't. Seperate from your or my claims, all motor textbooks I have seen give the power equation as:

IE = (I^2*Re) + [I*(E-I*Re)] at 227.4 Hz = 0.320. 0.227*1.41=(0.227^2*5.78)+[0.227*(1.41-0.227*5.78)] = 0.320

Balancing exactly note, and attesting to the accuracy of my measurements, which you unwittingly disparage. No need for phase to confirm power magnitudes in this case.

By the way, this dispells your manipulating my data to get I = 0.232. You need to face reality and use measured magnitudes, but I suppose manipulating suits your needs.

2) the omission of the rest of Rms in what you have done.

A taste of your own medicine.

Sly, but no cigar. It is found by your *manipulation* of my data. The actual measured magnitude with Fluke and HP meters is I = 0.227 amp.

I did address this, regardless... I'll use phase, if you have no problem with my result, then your Pmec of 0.0054 is wrong.

P = EI cos angle E = 1.41 I = 0.227

phase angle between voltage and current is arc cos Re/Ze = arc cos 0.931 = 21.44 degrees where Re = 5.78 and Ze = 6.21 (both *measured*, note. Note?)

P = EI cos angle = 1.41 * 0.227 * cos 21.44 = 0.298

Showing your phase angle and magnitude of 0.3168 watts to be wrong when correct measured magnitudes are used. This alone should make you take heed. Perhaps you are so out of sync, you do not realize this is the power lost as heat in the coil, leaving mechanical power to be

Pmec = Pin-Plost = 0.320-0.298 = 0.022 = I Eb

Do you have a problem with this approach, using phase? Again, if not, your Pmec of 0.0054 watt is wrong.

I have made appropriate corrections. You have not.

Read the above yet again: "You have been told multiple times that velocity was *measured*. " Drop the BS. However I guess you can't, as you depend on it for support.

Sly, but transparent. Again, your erroneous magnitude comes from YOUR MANIPULATION OF MY MEASURED DATA. Got it yet? Shall I repeat? As if I have to ask...

No. Absolutely not. It leaves 0.320-0.298 = 0.22 watt, as shown above. Sorry, but you are so hung up on your phase bit that you appear wigged out. Otherwise you don't know how to apply your theory.

Well sorry, I guess I overlooked the question out of courtsey in not responding as I felt, so I shall respond forthwith:

If you had the practical experience I've had with meters, you might not make the dumb-assed assessments you've made. Sorry, but on second thought, it likely wouldn't have helped you at all...

Like I said, it likely wouldn't have helped you at all.

?on trust as being correct, I am now finding somewhat

I gave you this information, and you did not contest it. Good thing we're not talking about displacement at 20,000 cycles instead of 227.4 cycles, otherwise you'd likely have displacement in the negative area.

I gave no 0.1 magnitude, that was more of your manipulation. By the way, have you ever even seen a Fluke, HP, or Simpson meter, let alone used one?

Just asking you to show the same courtesy I show you, but obviously courtsey is not in your bag of tools.

A thankless job, since when you are proven wrong, you just weasel out.

In most of your rambling, it's not worth the effort.

No, it is you having a problem with Rmec as given by Halliday, Serway, Beranek, Villchur, Colloms, and me. This was proven with their damping coefficient in forced oscillations of Rmec = 11.12.

show them unaltered and with magnitudes. Until you stop

Exactly. They all say the damping coefficient in forced oscillations at 227.4 Hz in this case is 11.12, not 2.23.

Avoiding? I'm here. Halliday, Serway, Beranek, Villchur, and Colloms are here. It's you who's out to lunch.

This is pure BS and is disingenious. The equations giving 0.027 watt are absolutely applicable and apply to forced oscillations. If you don't know this then you should be about finding out, otherwise stay out of what you are not qualified to discuss.

No, it is you who doesn't understand the dynamics of a reciprocation motor as well as you do rotary motors, plus you are hung up on *electrical* theory.

I told you CLEARLY several times, no point in getting into side and unrelated or minor issues until our difference on Rmec and Pmec was settled. In any event your questions here are now all addressed. Happy?

And I have shown you a more logical place it comes from which would have been the next and more interesting point to iron out.

Eb (Bl)/Re (Bl)^2/Re = --------------- = back resistance = 8.89 v

It is a resistance opposing the drive force. Whether it is due to motor resistance as Small indicates, or back emf, or resistance encountered in moving the mass, or some combination of these, or whatever, it is a resistance that must be overcome. As such it is included in the damping coefficient Rmec at 227.4 Hz. It does drop out of Rmec at resonance. I gave reasoning for this, but you rejected that reasoning in your strange manner of disregarding Halliday and Serway as well.

Then get Beraneks book. You should have had it before beginning a serious discussion on forced oscillations. Oh,I see.. Small needs Beraneks book, but you don't...

It's in Beranek's book. Just get the freakin' book and see his relation regarding force, velocity, and power, as derived using Rmec.

Get Smalls papers and you won't have to wonder. I dislike arguing with someone who wonders about the facts.

You feel no need to face the facts, that's your *real* need. So far as continuing the thread, you should have obtained the facts before jumping in.

Yes, take your errors into privacy. A little late for that, I would say. Bill W.

---------------------------------------------

---------------------------------------------

Reply to
Bill W.

--------- Actually, you forget that I repeatedly said that this damping component was

11.12 and used it in my calculations - Claiming that I did not accept it is a false canard. Note also that I suggested a way for you to arrive at this expression from F=[(Bl)^2/Re +Zm]V Its again apparent that you didn't see the relationship.

--------------

--------------- In what way. If there are errors in the manipulations- point them out. I am as capable of rearranging the power equations as the "experts" - it really is quite simple.

The only thing that I have not done is use your Rmec in power calculations and have repeatedly given you the reasons why. Note that {(Bl)^2/Re}V^2 results from Eb^/Re and (Bl)^2/Re drops out of I=(E-Eb)/Re which is independent of the mechanical system. No response to this. In addition, using the Kinsler form, referring things to the electrical side, the (Bl)^2/Re term is non-existent. When Kinsler writes P=I^2Ro +u^2(Rm +Rr) there is no term related to (Bl)^2/Re I've told you this before>

---------------------- Thank you for the correction - at least I didn't continue to use it or jump on cases of sloppiness on your part- recognising a typo for what it was.

-----------

On the basis of E being rms then F=BLE/Re =1.75 rms and has acorresponding peak value of 2.475 so what all the rest of the calculation appears to be an attempt to confuse. Are you telling me that the 1.75 was "average" because E=1.41V was also "average? . Discrepancies again- please note that I used, right from the start the value of E=1.41 V as being rms. This is something that I asked and you confirmed. Are you saying that you lied? All calculated values were based on this being true. If E was an average value (and voltmeters wouldn't normally read this on an AC range being generally calibrated in rms), then all numbers that I calculated except for power hold true as average values. The power will need a crrection of 1.232 so Pmec =0.0066 in that case

----------- I actually used P=(1.664root(2))^2 *2.23/33.82^2=0.0054 Using the actual force and actual Zm

---------------

-------------------- You have shown nothing new by this. All that you have done is used the value of 11.12 as the resistance vs the value of 2.23 that I used. Note that Halliday, Serway, etc do not say what you claim they said- Again something that was answered before- using their equations. you have not presented any evidence that they use this (Bl)^2/Re in calculation of mechanical power.

roup=alt.engineering.electrical&as_usubject=&as_uauthors=bill+w.&as_umsgid

--------------- KEY words "is included" I recognised that and also pointed out that you had not used it. You made the above statement but nowhere did you use the 0.629 value independently in any calculation.

------------ Not weaseling but simply pointing out a fact. You have presented absolutely no prior calculations with this value of 0.629 being used.

-------- No- I assumed it was a mistake but I see now that it was intellectual dishonesty.

------------>

------------------- Hardly

---------

----------------- The problem is that the (Bl)^2/Re term has absolutely no frequency dependent terms in it. it will remain at 8.89 at all frequencies. You do use it when you calculate the velocity at resonance.

1.75=11.12V giving V=0.1574 m/s You use it at 227.4 Hz in 1.75=root(11.12^2 +33.75^2) V to get V = 0.0493 At any other frequency say 100Hz 1.75 =root(11.12^2 +(628.3*0.0253 -3425/628.3) V to get V = 1.75/15.3=0.114 At f=59Hz 1.75 =root(11.12^2 +0.1398^2) V to get V =0.1573+ No transition there. Now look at power Frequency P your way Pmy way fc 2.23(0.1574)^2 =0.0552 watts 0.0552 watts (same) 59Hz 11.12(0.01574)^2 = 0.275 watts 0.05517 watts 100Hz 0.1445 watts 0.0290 watts 227.4 0.027 watts 0.0054 watts Try this as I asked at other frequencies (fc +/- a wee bit or well below resonance) There is no special transition formula needed- empirical or otherwise You are building in a complication where there is no need or justification for it. The 11.12 term remains at all frequencies - the damping factor that you so fervently remind me of. The actual mechanical resistance is still 2.23 Ns/m

---------------------

------------------------

Cute: Please look at the system that he refers to; It is the purely mechanical system as represented in fig 1.6 or as in fig 1.10and 1.13. There is no coil, no electrical source, no magnet, no Bl ,no current and the force considered is an externally applied mechanical one. It is not a loudspeaker. There is no (Bl)^2/Re term present. This is a red herring

----------

------------------- Ok you have plugged numbers into an equation without considering whether this equation, as is, is applicable to the loudspeaker. As I said above the force is of mechanical origin and the only damping term Rm is the mechanical resistance - not including any effects of a non-existent coil and electrical system . Certainly if you plug in a force and velocity as indicated the result will give you the value that you want- but it doesn't prove your contention in any way. F=2, V =0.5 then R =2/0.5 =4 fits the equation just as well. You seem to have a strange interpretation of this formula which is that of a time varying quantity v(t) =(Fmax/Rm} cos wot =(Fmax/Rm ) cos 368.2t =Vmax cos 368.2t This is a cosine function and will be Vmax at 368.2t =0, 2pi, etc

-Vmax at 368.4t=pi, 3pi, etc and 0 at pi/2, 3pi/2, 5pi/2 etc

note that at T/4 the value will be 0 as 368.2*0.00427 =1.57 =pi/2

------------------- Again you are misusing a formula to try to prove a point. This equation doesn't support your contention. Also you are conveniently ignoring what I have aways shown - that V=F/Z where the Z=11.12 at resonance (see above). The value I get for velocity is the same as you get. You apparently do not pay any attention to what I have said but go over the same old false accusations - see the data above. I do use the damping resistance of 11.12 because it includes the damping provided by the electrical side.

--------------------- Please note that none of the books use this equation look at it and what you have written is IE =I^Re +IE -I^Re which =IE Let I=2 and E=4 IE=2*4=8 I^2Re +I[I*(E-I*Re] =4Re +2(2*4 -2Re) =4Re +8 -4Re =8

You are quoting a power >

------------- Balancing because it is phoney. You appear to have distorted an equation for a DC motor EI =I^2 +EbI which is quite valid----FOR DC. Since the system is AC and there is reactance, the velocity and force are not in phase (else why use power factor) so the current and Eb are not in phase nor are the current and source voltage. Phase relationships are important. Your "exact" calculations are based on ingorance of the facts.

---------- Lets see: you gave me the values that I used. I did not distort them in any way. Again I gave all my work in detail so you could follow and pick at errors - no response. I note that as the data used is based on your experimental results and each value can have errors, the correlation between

0.227 and 0.232A is within 2.2% which is really a credit to your measurements. I also have pointed out that the measured current includes factors such as coil inductance which are not included in the data or the model.

-------------------

--------------- I did not manipulate your data- if you see any manipulation- it was all there. It is interesting that playing with L does affect the calculated current. Again, which side of the ammeter was the voltmeter? Did you measure the current with the supply at 1.41V or did you measure it with the coil side of the ammeter at 1.41V. What was the vlotage drop across the ammeter.

-------------

------------- Your calculated p=0.298 is the input power. That is correct. The value of 0.320 is the VA input. (EI) That is 0.320 cos 21.44 =0.298 watts You are comparing VA to Power - fundamental error. You seem to be geting into more and more of these unthinking and incorrect calculations.

--------- I did it correctly the first time.

--------------

--------- Show where manipulation occurred. I have aske for this earlier and got no response.

---------------

--------- see above

0.320 is the VA, the hypotenuse of the power triangle. The base is 0.298 - the power component. Sorry, you have failed elementary circuits. Also , your 0.22 watts is a far cry from your 0.027 watts.(11.12* 0.0493^2) Howcome?

----------- you haven't answered the question I've had as much practical experience with meters as you have - different types, different measurements and in a variety of situations -that is why I question measurements.

------------- I don't recall you giving this information regarding the actual measurement technique. The question above has a catch if you don't realise it.

------------ No- you gave nothing- that is why I asked. Yes I have, as well as power factor meters, wattmeters, precisionbridges and meter calibrating equipment. I have also used scopes, counters, etc. I have also used brands other than the above. >

--------- It is not a courtesy to plug numbers into an equation and then say that it proves your point. I could write E=ReI +BlV as 1.41 =5.78I +7.17V I could also write 7.17I =2.23+j33.75)V and have two equations in two unknowns.I and V both of which will have phase angles, for which I have no values (oh yes -i could plug in your measured values but then I would have two equations that don't make sense as you didn't measure phase angle information.) Therefore I do what Kinsler, Berenak, et al do- I re-arrange the equations into a more convenient form before plugging in numbers and calculating magnitudes and phase angles. This I did and laid out for you, step by step.

-------------- You havent come near proving me wrong.

-----------------

----------- Or you cannot follow it.

------------

------------- I have no problem with their work and their dampi9ng coefficient (11.12 is yours and I agree with that. I REPEAT - I use this term in calculation - it hangs right out there where you can see it. As part of this is a reflection of the effect of the electrical resistance on damping, I omit that part from the mechanical resistance in calcualting mechanical power. Results are consistent, power balances are consistent and checks by using Kinsler's models result agree.

-------------- While they are all saying the same thing the equations in ch1 do not deal with the situation where there is an Re or Bl or E - they deal with a strictly mechanical system, not a loudspeaker. They cetainly do not support your point as you think.

--------------- They are there- try to read them for understanding. They are fine. They had good teachers.

---------- The equations don't give 0.027 watts. They will give an answer based on what you put in them. If you put in nonsense, then they will give nonsense. That you have done.

------------ There is no fundamental difference between reciprocating motors and rotational motors. Feed low frequency AC to a DC motor and it will oscillate. As far as electrical theory is involved- the models used by berenak,Kinsler and others are all equivalent electrical circuit models. Electrical theory is the tool used for the analysis of these models.

--------------- No, I am not - the answers are even more erratic than before and the equations and steps that I have given are not side issues as they deal directly with the difference on Rmec and Pmec.

------------- Again, I include it in damping and always have done. However you have continually and deliberately ignored that. Actually my point regarding this is suppported by what you say above. Small attributes it to motor resistance and ,yes, it must be overcome. Yes, (Bl)^2/Re =Eb/Re if that is what the messed up equation above implies. It is the component of the damping due to the coupled electrical circuit resistance and thus is involved in the F/V ratio. As its origin is in the electrical part of the system, it doesn't contribute to the power on the mechanical side What's the use.

--------------- No, forced oscillations in a mechnical system and forced oscillations in an electrical system are essentially the same. I have a sufficient background in that - .

-----

--------- Does he do this as Kinsler does, in an introductory chapter dealing only with mechanical systems. - the equations you misapplied to try and prove your point?

------------

-------- Actually to make it less embarrassing for you. You haven't proven a thing but have gone around in circles repeating the same things and not backing any of it up. You haven't given a good answer to any question that I had posed and avoided critical questions. When you do give an answer, it falls apart on cursory examination.

bye

Reply to
Don Kelly

You made false claims and said "Bye". No fair.

Reply to
Bill W.

---------------- Again and again, how many times do I have to tell you. I have never said that the damping coefficient for this loudspeaker was 2.23. I have always used 11.12. What I have said is that the electrical resistance contributes to the damping. You are beating a dead horse - can you read? Where I differ from you is that I do NOT include power in Re ( or its mechanical equivalent (Bl)^2/Re )in the actual mechanical resistance for power calculations. Neither does Kinsler. In fact there are at least two ways to represent the speaker and its damping where the term (Bl)^2/Re doesn't appear. Kinsler shows one.

-------------- If you had read what I said, I noted that the whole of chapter 1 is a review of basic mechanical systems. . The references above do mention loudspeakers but only as examples of mechanical vibrators. I suggest that you look at the context of the three references above. The material of the chapter deals with purely mechanical systems, not electromechanical systems. Certainly mechanical principles apply to loudspeakers. However, to use Ch1 to support your contention regarding the (Bl)^2/Re factor due to the electrical parameters is either dishonesty or stupidity. As I pointed out- the F=BlE/Re term and (Bl)^2/Re term simply do not exist in any of the material of chapter 1. Loudspeakers are electromechanical systems- in ch.1 there are no electrical terms or couplings to electrical systems.

-----

--------------- A problem? Re-arranging doesn't properly doesn't change relationships. Any steps in re-arrangement have been shown.

----------

No- it means that I had a typo which should have used [(Bl)^2/Re]V =Eb/Re

-------------

------------- But as I pointed out - this is really a "re-arrangement" of BlE/Re =root[9(Bl)^2/Re +Rms)^2 +(wm-K/w)^2]V I have no problem with this -reaosn given many times before. How is it that you have no problem with such a "re-arrangement"?

Reply to
Don Kelly

You are going in circles, as we have discussed this. Until you understand that power is required to overcome the (Bl)^2/Re motor mechanical resistance or perhaps more aptly put, the motor's equivilant mechanical resistance, and that (Bl)^2/Re must be included in the power equation as a resistance to be overcome, and that this requires power dissipation, your mechanical power magnitude of 0.0054 watt will be incorrect.

Physic texts almost universally use a mass on a spring for unpowered examples. Kinsler speaks of power supplied to the system in the sentence before his first power equation, between the first and second power equation, and between the fifth and sixth power equation (page 14). He uses a loudspeaker as example since it has a motor to supply power to the vibrating system. You know... a motor to push the cone. Sorry, but perhaps it is you who should read. You think?

Then those copulating mice of yours are at it again.. Rats.

Certainly

I am honest, smart, and benevolent, thank you. Please take note:

I'm a lover of peace, protector of the downtrodden, (always for the underdog, unless he's losin'...)... I'm true-blue Oatie Collotie.. Descended from Royalty. Constant as the north star - There when you need me... With words that serve to inspire and enlighten... Peace unto you Brother, and the horse you rode in on.

As I pointed out- the F=BlE/Re

What is the phase angle for Eb that you use here? TIA

Is V volts or velocity as used here? TIA

It is the magnitude of mechanical resistance, as taken from the solution to the equation of motion by Halliday and Serway. It is correct, Rmec = 11.12 at 227.4 Hz.

I assume the 9 is another typo?

Why do you give the force equation as a "re-arrangement", and what is the significance of doing such? TIA

You know very well that it shows mechanical resistance Rm by Kinsler to be 11.12 and not your 2.23, giving mechanical power in equation 1.35 at least five times greater than your magnitude of 0.0054 watt.

By the way, I checked Small, and he defines Rmec as Rmec = (Bl)^2 + Rms, as well as Beranek (eq. 8.)

That Kinsler's Rm is 11.12. You know this. Too bad.

No more than my use of it.

Prove it please. TIA

We shall see. I gave my sulution, so give your solution to the equation please. Here it is again for your convenience. TIA

F Rm = --------------- cos wo t = 11.13 v resonance

---------

It's about time you recognized my most superior intellect. Now the next step please, is for you to appreciate it. A bow may be in order.

A non-sequitur.

Bill W.

Reply to
Bill W.

"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...

contributes

---------- No, I am not going in circles. The (Bl)^2/Re is an artifact of the model - It simply comes from the equation E-RI=Eb No more than that. The "power" in this term is neither the I^2R term nor an actual mechanical loss term. I have told you this before and detailed where and how this term arises. Please go to a decent circuit model text and study Thevenin and Norton equivalents - the information is there. The fact that, using the same data and a different approach gives the same result should mean something. Note that in Chapter 14. the Rm that Kinsler uses is the actual mechanical resistance (and nowhere does he use this term as anything else. Certainly the effect of Re must be included in the damping (either as is or as (Bl)^2/Re depending on the approach used) because the electrical and mechanical sides of the speaker are coupled.

-------

loudspeakers

--------------- Elementary physics texts do so. Better texts or texts on dynamics as used in freshman courses do cover mass, spring damper systems with external forces as shown for the simplest case in the freebody diagram of Fig.1.6. Note that chapter 1 of Kinsler deals strictly with mechanical systems. Mass, spring, damper and a mechanical force. The source of the force is unspecified. See section 1.7 and read it carefully. The equations of ch.1 do apply to the mechanical half of a loudspeaker, which, without the electrical side, is no more than a trampoline for mice. What makes it a loudspeaker is the electrical /mechanical coupling- i.e. a motor if you wish. This coupling which makes the motor isn't dealt with in chapter 1 which deals only with the "load" on the motor (including its mass, friction, etc as well as the useful output) It is only in chapter 14 that Kinsler deals with loudspeakers per se. Note the definition of force therein as well as the use of force in Eq 14.23. In that chapter he allows for both electrical and mechanical input. Note also that in chapter 14, the impedance Zmo =Rm +j(wm-s/w) is the open circuit mechanical impedance- Open circuit in that I=0 (or the equivalent, Re=infinity) for which BlE/Re =(Bl)^2/Re =0, zilch, nada. In that case nothing on the electrical side is reflected to the mechanical side, either force or damping. Then all that is left is Rm which is the suspension resistance (Rms-Rr), the mass and the spring. Then the power expression that he uses just after Eq14.53 indicates a total power input of ReI^2 +RmsV^2 which doesn't include any [(Bl)^2/Re]V^2 which is, as I said, ad nauseum, an artifact of the model used.

---------- Only for large woofers.:)

------------

-------- same as that for V

----------

--------- as used throughout to date- velocity. Kinsler uses a bold faced u to indicate a phasor velocity. I have used E for voltage. (v ,e, i, f for instantaneous values). I would prefer to use a bold faced E,V etc but newsgroups still prefer to use plain text. I sometimes use |V| to indicate a magnitude.

----------

---------------------- However, it is still a re-arrangement of the equation as I gave it above. The equation is not a solution to the equation of motion but is, in fact simply a re-formulation or re-arrangement of the equation. The solution to the equation depends on what values you put into it. You have given an answer but without indicating the values that you used as input. I could use other values to get a different result. The equation doesn't care. NB. Fmax/xmax =Frms/xrms=Fave/xave =Fw/V = w*|Z| ( max, rms or ave as long as both are in the same form-in this case it is w*|Z| = w*35.54 (m/w)(w^2-wo^2)= w(wM-K/w) =w*33.75 Halliday and Serwick's equation with these numbers supplied, simply becomes b=root {[(w*35.54)^2-(w*33.75)^2]/w^2 =root [35.54^2-33.75^2]=11.14 with round off. What's the big deal? There is a problem if it gives b=root (negative number) which indicates an error- a physically impossible situation. In addition, the form of this equation exaggerates computational errors.

---------- Definitely so, sorry.

-------------------

----------------- I said that the equation for b as given by H and S is a re-arrangement of the force equation (which itself is one re-arrangement of the basic electrical and mechanical equations). The only reason for re-arranging an equation is to put it in a form which is more suitable for the task at hand. If ,for example, you know the force and displacement at a frequency w as well as mass and the resonant frequency then the Halliday and Serwick form is convenient for determining the damping factor. If you know the damping factor already then this form is really not worth bothering with. Note that the equations of motion in chapter 14 are expressed in "re-arrangements" of Eq 14.2 to get a more convenient form for calculation.

-------------

----------- No it shows that 1.75/0.157 =11.13 I could use 0.350/0.157 = 2.23 just as easily and just as correctly. Noting that this particular value is at resonance, don't you see a discrepancy with your own calculation that the power is 2.23(0.157)^2 =0.055 watts? My contention is that there is nothing different at any other frequency except that the velocity changes so the mech power is still 2.23(V^2) which is in line with Kinsler.

--------------- I have no problem with this as long as the Rmec is used for calculation of velocity from the locked cone force (BlE/Re). It does represent the effect of the electrical resistance on the damping. However, the key is- do Beranek and Small use this value of Rmec to determine mechanical power? What happens to I^2Re ?

-----------

----------- See above. You have 11.12 because of the values you put into Kinsler's equation -NOT because Kinsler's equation uses these values. Kinsler's equation gives an answer for whatever numbers you plug into it, right or wrong.

--------------

----------- In normal usage of f(t) =Fcos(wt), the cos(wt) implies a cosine wave (a sinusoid which has its peak at t=0) That is f(t)=F cos 2*pi*f*t =Fcos (2*pi*t/T) At t=T/4 this becomes Fcos(pi/2) =0 N.B. the angle is in radians not degrees (2*pi) radians =360 degrees. This is normal usage in all texts. It should be explicitly dealt with in any circuits text.

Secondly, pf has no meaning at a given instant of time. It is really a factor related to continuous steady state sinusoidal operation- normally expressed in terms or rms values. In addition there is nothing special about t=1/4T as opposed to any other time.

---------

You have misread Kinsler's equation. If you go back a bit he uses f=Fcos(wt) - a time varying force and the impedance reduces to Rm so that f=Rmv becomes Fcos(wot)=Rmv so v=(F/Rm) cos(wot) - also time varying and in phase with the force. this leads to: Rm =F/ V

Rm is not time varying, nor is cos(wot) a power factor. I had already given an alternative solution and repeated it above. You confuse the equation with the numbers that you plug into it.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

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