Motor torque and back emf

So do Halliday and Serway.

You need to look at Small and Beranek. Why do you not do this? Small had produced by *far* the most profound work on speakers in the last 50 years, and he references to Beranek *many* times and *none* to Kinsler. Again, why do you not get the Book? Want me to get one for you as a gift?

[(Bl)^2/Re]V does not equal Eb/Re. You are wrong, therefore don't understand how to use (Bl)^2/Re.

Halliday's mec resistance equation doesn't need rearrangement. You need to rearrange your ploys.

I disagree, and have shown why.

I have stated what happens to I^2Re before. To restate: Power out equals power in minus losses. In this case the loss is I^2*Re, so that Pout = Pin - loss = Pmec = 0.022 watt. Note IE cos angle = IE Re/Ze = 0.320*5.78/6.21 = = 0.298 or I^2Re = Power loss in this case. Then 0.320-0.298=0.022. Phase taken care of. Your error re Pmec = 0.0054 watt taken care of. You have tried to refute this by not using my measured magnitudes, then hooting about your calculations being based on my supplied values. This is now to a point that you are either ignorant or disingenious.

Pmec = 11.12.

Bill W

Reply to
Bill W.
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-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer "Bill W." wrote in message news: snipped-for-privacy@news.supernews.com...

------------ Good for you - you caught that. you are absolutely right - try [(Bl)^2/Re]V^2 =Eb^2/Re which is correct but doesn't mean anything valid. .-------------

calculation.

----------- I didn't say it needed rearranging- can you not read. I said that it IS a re-arrangement of the force equation and no more than that.

----------.

---------- No you haven't shown why. You have simply plugged numbers without justification of the numbers that you plugged.

------------- But my contention was that E and I are not in phase so EI is NOT power but VA- hypotenuse rather than base of the power triangle. EI-EI*pf as you have shown, is garbage. Using P=EI*pf is valid, provided that you use the correct pf, which you haven't.

----------

-------- Several things:

1) magnitudes alone don't tell the story. 2)You use IE( Re/Ze) which assumes that the power factor is Re/Ze. That implies that Ze is purely reactive- no resistive component- and no associated power component. This implies that there are no mechanical losses. You have noted that I^2Re =0.298 =IE (Re/Ze) which also implies that all the input power is lost in I^Re leaving nothing for mechanical power. This also says that there is no mechanical power. 3)You have attempted to get around this by using IE as input power where it is not. E and I are not in phase. This is based on wishful thinking. If IE is the power then E and I are in phase. If E and I are in phase, then how can you have a power factor which is not 1? This you have done. I don't think that you are trying to bamboozle me- you have bamboozled yourself. 4)Now you get Pmech = 0.022 which is about 23% LOWER than the 0.027 value you calculated before( which on the basis of velocity being average, is itself about 23% lower than the "actual" power-net result about 51% lower than "actual". Perpetual motion machine maybe? Don't these little inconsistencies bother you? As to the data that I used in calculations- I laid it out exactly what I used. I assumed that these values were based on measurements that you have taken. You know how you got them. I note that you have not given an analysis, starting from the mechanical force and velocity and going back to find the current and various losses, which is free from discrepancies galore. I have, at least, got consistent results and any and all errors that exist are in the data you gave me. Here is what I used as a base: E=1.41V (assumed rms in absense of other information and the fact that your meters particularly the Fluke, measure rms) M=0.0253 Kg K=1/s=3425 Rms=2.23 Re=5.78 Bl=7.17 You make a great fuss about the current not being exactly what you measured. Considering all the accumulated errors that are in the data, the results are damned good. A 2% difference in current magnitude considering the possible cumulative errors in all of the data items above is excellent and a credit to your measurements. 5 to 6% could be expected. There is also the problem of rms vs average which is still not resolved- this puts a question mark on the value of Bl. (another inconsistency). Are you telling me that your Fluke gives AC voltage and current values in terms of average values?- not according to their web site. The Simpson will be calibrated in rms for a sinusoid while the Fluke is likely true rms. Do you know what the burden of a typical Fluke meter is and what its effect on measurements? Note also that a typical Fluke will have an error margin of 1.5% of full scale +3 digits on AC current- this in itself exceeds the 2.2% difference. There is a difference between meter precision and the accuracy of measurement. You can get the wrong measurement to great precision because the meter precision is not the whole picture. Frankly, I suggest that any inconsistencies are yours. You have presented a great deal of ignorance as far as the basic principles are concerned. You have not presented any good arguments. This topic is so bloody simple that it is not necessary to go back to "xxx says' (particularly as you have avoided the question of what they actually say with regard to power). Judging by some of your misinterpretations of equations given by others as well as what I have said, I have very little faith in your understanding of the physics, the math and even ...... Oh, Hell, why go on. It is hopeless.

C'est tout.

. >

Reply to
Don Kelly

Lack of response to a solution noted.

Nit picking noted.

PF = cos angle = Re/Ze = 5.78/6.21 = 0.931 , then IE*PF = 0.320*0.931 = 0.298 as noted above. Apparently you didn't recognize Re/Ze as power factor.

Aye Captain, you're the Dude with insufficient power. :)

Who me? Bamboozled? Naw... I told you...

E and I are at 21.44 degrees. arc cos Re/Ze = arc cos 5.78/6.21 = 21.44 degrees PF = cos angle = 0.931

Yes, the difference in 0.027 and 0.022 watt is worthy of concern. That was my preferred next step for discussion. However, in the power equation of power out = power in, 0.027 gives only 1.56% error. Your magnitude of 0.0054 gives 5.31% error. 0.022 balances the equation exact at 0.320 watt note.

No, the current I measured is accurate. I question you using a current magnitude other than that measured.

Yes, that was to be resolved, a difference of 10% in some parameters.

(another inconsistency). Are you telling me that your Fluke

Sorry, but what you have presented goes beyond ignorance. You are not willing to disolve your ignorance by checking where Beranek and Small include (Bl)^2/Re in the resistive term, i.e. as part of the mechanical resistance. Since they do so, I consider you in error.

Unfortunate that you see it that way, but I have told you repeatedly, I didn't come here to argue, but to learn and hopefully to contribute.

We discussed the power equation and failed to agree.

Your lack of faith im me means nothing until you get Beranek and Small's work, and yes, you are correct that it is hopeless until you do. My offer to get Beranek for you as a gift was sincere. The offer still stands.

Thank you for your time, and I shall now depart into the sunset...

Bill W.

Reply to
Bill W.

---------- If you wish. I'm not in a location where either are handy. Do you really think that will help your case, judging from the lack of understanding of the equations that you have quoted. I really doubt whether they said some of the things that you claim they do. I have no problem with Rmec as you state it for damping calcuations. Where we disagree is that I see it as an artifact of the model and based on the electrical resistance as seen from the mechanical side and as such, is not an actual mechanical resistance leading to actual mechanical losses . Kinsler shares this opinion and I doubt very much whether Beranek or Small fail to do so as well. It doesn't take expertise in loudspeakers to see this. In the area of loudspeakers and acoustics, I defer to the above. In the area of the modelling at the level we have been doing, I will not defer to them, without a far better argument than you have presented, because I have comparable knowledge of electromechanical systems. --------

------------- I see, it is nit picking to point out fundamental errors but it is OK to make them. If EI*(Re/Ze) =EI*pf is power then EI by itself is not. It seems that this "nit picking" destroys your argument. See your Siskind reference.

----------

------- I recongnised it -see below>

----------- Ok then this means that Z=6.21 @-21.44 degrees =5.78-j2.27 ohms. Then the reflection of the mechanical impedance is Z-Re = 0-j2.27 ohms PURELY REACTIVE This leads to a mechanical impedance of magnitude (7.17)^2/2.27 =22.6 ohms PURELY REACTIVE NO resistive element =NO power. It also leads to velocity =0.072 m/s (at angle 68.6 degrees or 90 degrees out of phase with the force BlI ) from (Bl)(e-ReI) among other absurdities. ---------

---------- You are still assuming the power in is 0.320 watts, which it isn't- but you admit to a power factor- E and I not in phase. EI is not power. P=EI cos(angle between E and I) =0.320*0.931 =0.298 is the input power. See the similarity to FVcos(angle) -if not see Siskind. Then I^2Re =0.298 from an independent calculation. Pin-I^2Re = 0 = Pmech ?????????OOPS?????

As for error 0.027 out vs 0.022 in leads to an error of 23% not 1.56% The perpetual motion people would love you. The 5% is expected in such a check calculation which is the difference between 2 nearly equal numbers of at best, 3 digit accuracy.

------------ In calculating the current I used the voltage you gave me, the parameters you gave me and came out with a current magnitude which is close to the value you measured. Any difference is due to accumulated errors in the data which you gave me. In case you don't realise it, this is a credit to your measurements- a difference which is really within the error band of the meter which at 1.5% of full scale on a 400ma range is +/- 6ma (=/- 3 digits in display. Did you take into account the meter burden? As to your power factor- that is a guess and a wrong one at that.

Reply to
Don Kelly

Don't judge me by your standards, Cowboy. I tried to wind this up amicablly, but no............

And you are not about to find out, preferring a dick-measuring contest. Problem is, you're not well equipped.

(Bl)^2 Beranek, eq.7.2: Rmec= ----------- + Rms + 2Rmr Rg+Re

(Bl)^2 Small, eq. 8 (June 72, AES): Rat = Ras + ------------- (Rg+Re)*Sd^2

Convert Small to mechanical with Sd^2

Do you see (Bl)^2/Re there in the resistive term?

Do you not understand resistance dissipates power, or do you prefer to hoot about wanting to see where Small or Beranek use the resistive term to show mechanical power? In any event I'll get to that below.

Do you not realize one of my goals was to define (Bl)^2/Re and its specific function?

Do you not realize if Beranek and Small had done such more specifically, there would be no interest for such?

Do you not realize your ass-off attitude thwarts one's efforts?

Do you not realize both of us could be in error in various aspects?

Do you not realize there is no need for your long show-off equation tirades in a simple aspect of discussion?

Do you not realize I came here for input on several parameters, and might be able to contribute as well, as opposed to coming here for a dick-measuring contest?

Do you not realize the helpful input offered by Daystrom in a decent manner is the best road to take?

Do you not realize assing-off generally gets the same returned to you.

Reply to
Bill W.

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

---------------- Again, you have not proven your point. This "Rmech" is used to determine the force/velocity relationship. I agree with that. If you had bothered to try to understand what I said, you would also see that. If you had bothered to try to analyse the original equations, you would also see that the (Bl)^2/Re term is an artifdact of the model used. I have shown this to you before - no response. Is ignorance bliss?

------

-------------- I found that out the first time I looked at the basic equations. How it arises is very obvious. Note that I showed you that it came from I=(E-Eb)/Re without any reference to the mechanical impedance. While it is useful for calculating the velocity, it is meaningless for power calculations.

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

---------------- Oh my. Do you realise that discussion which might be useful does not include dumb agreement? I have asked for criticism based on what I have written - not a quoting of generic equations which neither prove nor disprove any points. I don't mind being shown wrong- but I do need to be shown a rational argument- not a collection of simplistic statements which have no backing. ( Quoting P=F*V*cos angle and plugging in numbers does not prove your point)- particularly where you often give statements which fly in the face of physics and /or common sense.

----------------------- Why is it different in this case? You have made a claim- please back it up.

----------------- It is the part of the input or applied

---------------- Exactly

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

------------ Are you assuming E and I are in phase? Why and how can you assume this? E/I =Z so this implies that Z is purely resistive (pf =1). However, you do not assume this and have a pf of 0.931. Inconsistencies- at least you should be able to agree with yourself.

--------- One error is that you are taking EI as power. IT IS NOT unless voltage and current are in phase. In this case they are not in phase. Second error- the power factor as you have calculated. The power factor is based on the total resistance/total impedance. You have assumed the total resistance is 5.78 ohms =Re and that implies that there is no mechanical resistance. That is why EI cos angle =I^2Re giving a result based on YOUR assumed pf is that Pmec =0. This is obviously not so - A glaring discrepancy in your work. You simply did not read what I said and certainly did not check your assumptions. You have two wrongs which don't make a right.

-----------

------------ Actually, I have solved the system in different ways, without violating any of the relationships involved, and arrived at the same answer each time- showing all the work. You have not done this nor have you shown any consistency in your results.

---------

------------------ I have no problem with this - He is right.

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

------------- He doesn't use this- these are YOUR numbers and YOUR interpretation. I have pointed this out to you before as well as noting that the basic equation works quite well with the F and V that I use. I note that the F you use is that at v=0 - not the actual F at velocity =0.0493. You have also indicated that. That is a slight problem. If this is the actual F then the velocity must be

0 and the Pmech is then also 0.

If you used the actual Force (=BlI) and used the phase difference between this actual Force and the velocity, you get my answer from Berenak's equation. In addition, it makes physical sense.

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

---------------- You say this. Does Berenak? Did he know your numbers when he wrote the book? You have repeatedly used rather generic equations with a given set of numbers to "prove" your point, where the same equations with a different set of numbers gives, equally correctly numerically , a different solution. You have to use numbers which make physical sense. You haven't.

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

------------ As used with the force BlE/Re to find V - as I have done. Again, part of the modelling of the electrical admittance of a Norton source as transferred to the mechanical side. You haven't attempted to look at this, have you?

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

----------- Your interpretation - I disagree- Again, you haven't come up with a definite unambiguous statement from Berenak. I note that Kinsler does at least give an unambiguous statement. but of course, he now appears to be wrong, in spite of his modelling and physics making complete sense.

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

--------- Where proven? Not by you.

--------------------- when

------------------ That I have agreed with right from the beginning in the sense that the Rmec includes the effect of the epectrical resistance in the damping.

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

------------- You say that but you haven't shown that Berenak says that. Again, I very much doubt whether he would make such a fundamental error. Does he even bother finding the mechanical power or does he simply find the acoustic power and possibly the overall efficiency?

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

-------- Nope - I see that it is the result of 2 errors compared to another error

-----------

----------- I'm sorry to have upset you and admit that I should have been more polite. All I really want is not a pissing contest but to have you really think. I failed in that. It appears that I should not have questioned you on what appeared to be glaring errors. I expected a good defence of your position and some rational arguments and explanations of your position. I was wrong in that expectation. It really is too bad.

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

Reply to
Don Kelly

Energy input = energy lost in conversion + energy output

i.e.

Work in = work lost creating heat + work out

rearranging, then work per cycle at 227.4 Hz is:

work out = work in - work lost work out = (IE t)-(I^2 Re t) = 0.001408 - 0.00131 = 0.000098

I=0.227 (measured), E=1.41 (measured), t=0.004397, Re=5.78 (measured)

giving power = work/time of:

P = W/t = 0.000098/0.0044 = 0.022 watt

Proving your mechanical power magnitude of 0.0054 watt is incorrect, as well as your analysis. Your problem is you don't understand the power distribution on the load side.

Now..... which will look worse for you, trying to refute the above or admitting your error? If you cannot see the latter would make you look better a hundred times over, then our discussion needs to end, as I have no respect for someone who cannot admit an error.

Beranek and Small prove (Bl)^2/Re is resistive, and resistance dissipates power. They prove my point.

This "Rmech" is used to determine the

The only artifact is in your imagination. Ignorance?

Avoidance of answering noted.

Absolutely not.

Avoidance of answering noted.

Avoidance of answering noted.

Avoidance of answering noted.

Avoidance of answering noted.

Avoidance of answering noted.

Avoidance of answering noted.

It includes misunderstanding on your part.

I have asked for criticism based on what I have written -

I don't mind being shown wrong- but I do need to be shown a rational

I have worked my way through *both* our errors to a successful analysis. That was what I came here for.

The answer follows next that you agree with. ????????

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

Phase is your problem, not mine, not Beraneks, not Smalls, not Fitzgeralds, and I agree with myself and these guys as well.

Your judgement on this is unworthy unless you can refute the conservation of energy equations top-posted.

My consistency varied on the way to finding solutions. Your consistency consists of hanging on to your incorrect power and resistance magnitudes.

My magnitudes make sense. See Fitzgerald et al top-posted.

See top-posting about who makes sense.

You based your claim re being correct on Kinsler, and now state he appears to be wrong? In what way do you now see he might be wrong?

Yes by me. It's in the last few posts. You apparently have a problem either with memory or with being disingenuous.

Reistance cannot damp without dissipating power.

There is no error.

Does he even

No error, See above.

I'm not upset, nor am I the one taking several days to respond. You give yourself too much credit with the ability to upset. Kindly note that

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

Your lack of understanding speaks for itself, and that is what is too bad.

Bill W.

Reply to
Bill W.

------------ Actually you have dug yourself a deeper hole. You have proved your own lack of understanding of fundamentals. Your energy IEt is correct ONLY if I and E are constant over the time t. They are not. As a function of time, i(t)=I*cos(wt+phase) and e(t)=E*cos(wt) (wt is in radians, not degrees) (here I and E are maximum values) The product or instantaneous power is E*I*cos(wt)*cos(wt+phase)=E*I*[cos^2(wt)*cos(phase) -cos(wt)sin(wt)sin{phase )] Integrating over the cycle gives W=integral from 0 to 2pi radians of [E*I*cos(phase) ] =0.002816 cos(phase) (integrating over the quarter cycle will not be correct except at a phase angle of 0-try plotting the energy waveform to see that- the energy waveform is at 2*277.4 Hz). Dividing by the period gives Pave=0.5E(E*I)cos (phase) =0.032 cos (phase) as the average power In terms of rms quantities this is Erms*Irms*cos (phase)=0.320cos(phase) Kinsler does this (Eq.1.34) for force and velocity while Siskind does itfor electrical circuits. Fitzgerald will also do this or assume that the reader already knows this. Go to any applicable text to find this out. It also appears that you are still using DC analysis (as for the DC motor) for AC conditions- This is not valid and you will see that the authors cited will, in fact consider phase for AC..

This is definitely NOT the same as the result of multiplying the average E times the average I times the 1/4 cycle time (as you have done)which is incorrect except for a square wave or DC. This is a basic error (and one you have fallen into earlier).

If the phase angle is 0 then the mechanical impedance reflected to the electrical side is purely resistive (not true)-and appears as 6.21-5.78 =0.43 ohms - reflected to the mechanical side this is (7.17)^2/0.43 =119.6 Ns/m purely resistive! Can't be- so quite possibly E and I are not in phase. This assumption regarding angle is the one which you have used to find Pmec=0.022- lets use that. Then Rmec =0.022/0.0492^2 =9.09 which is less than 11.12 { (Bl)^2/Re =6.86 so Re =7.5 ohms which is not what we started with and is greater than Z=6.21} Again a puzzlement. Your Pin of 0.320 actually leads to two impossible answers.

Your arguments are getting more and more frenetic and more and more incorrect. Please re-read, carefully what the various authors are saying and do bone up on AC analysis as well as a very useful tool-calculus. Your authors use both correctly. I can't be bothered any more.

Old crap and snide comments snipped.

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

Reply to
Don Kelly

" Integrating over the cycle gives W=integral from 0 to 2pi radians of [E*I*cos(phase) ] =0.002816 cos(phase) (integrating over the quarter cycle will not be correct except at a phase angle of 0-try plotting the energy waveform to see that- the energy waveform is at 2*277.4 Hz). Dividing by the period gives Pave=0.5E(E*I)cos (phase) =0.032 cos (phase) as the average power"

This should be:

Integrating over the cycle gives W=integral from 0 to 2pi radians of {E*I*cos^2(wt) *cos (phase) } =Pi* E*I *cos (phase) =2.0 cos (phase) Joules

(The integral of cos(wt)*sin(wt) =0 so is disregarded.)

(integrating over the quarter cycle will not be correct except at a phase angle of 0-try plotting the energy waveform to see that- the energy waveform is at 2*277.4 Hz).

Dividing by the period (2pi radians) gives Pave =(EI/2) cos (phase)=0.320 cos (phase) Watts

Sorry for the errors in trying to short cut from your numbers. .

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

e(t)=E*cos(wt)

E*I*cos(wt)*cos(wt+phase)=E*I*[cos^2(wt)*cos(phase) -cos(wt)sin(wt)sin{phase

Reply to
Don Kelly

I understand the fundamentals of conservation of energy alright. It is you trying to slip around it and offering Fitzgerald as support. No point in going into details, as this speaks to your confusion, if not denial of reality. Then you again launch a tirade of equations, supporting your view, and say: "This assumption regarding angle is the one which you have used to find Pmec=0.022- lets use that. Then Rmec = 0.022/0.0492^2 =9.09 which is less than 11.12 { (Bl)^2/Re =6.86 so Re =7.5 ohms which is not what we started with and is greater than Z=6.21} Again a puzzlement. Your Pin of 0.320 actually leads to two impossible answers."

Yes it was a puzzlement, and I understand your hanging onto your equations. I solved the puzzlement, you haven't. I did inform you that a modifying term was needed and described such, but you went into yet another round of equations, and apparently it flew over your head. It all ties in with what (Bl)^2/Re actually represents, which I was in error about early on. What I fail to understand is that when Beranek says Pmec = force times the in-phase component of velocity and this gives 4 or 5 times your magnitude, why you do not take pause. After all, Beranek wrote the Bible on basics and Small wrote it on application, noting reference to Beranek six times in the first two pages of his analysis. Nor do I understand why when the conservation of energy method gives the same results of 4 or 5 times your magnitude of 0.0054 watt, why you do not take pause. I can only assume I have overestimated you, both in the math department and whatever department where it says a man should be responsible for his mistakes. Also I note you did not answer my question re Kinsler in my last post, which might have led to some agreement, so this supports my view.

So far as you being bothered, I'm sorry about that, but I feel you are stuck in a rut of your own making.

Bill W.

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Reply to
Bill W.

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