Looking for Forums for Electronics Engineering Design, etc.

How much do you pay. . . I DO NOT FOLLOW MANY OF THESE NEWS GROUPS To answere me address mail to snipped-for-privacy@aol.com

----- Original Message ----- From: "Bill W." Newsgroups: alt.engineering.electrical Sent: Thursday, December 11, 2003 11:15 AM Subject: Re: Motor torque and back emf

--------- Actually the magnitudes of the terms are given in my original calculation. I showed the polar form which is (magnitude) @ (angle) as well as rectangular form (real) +j(reactive)

For data I used only the values of E=1.41 volts, Re=5.78 ohms, Bl=7.17 Newtons/Ampere or volt-sec/meter Rms=2.23 N-s/m, M=0.0253Kg and K=3425 N/m as I had indicated.

The following were calculated. V=0.0492 @71.76 degrees (magnitude 0.0492 m/s Zm =33.82 @ -71.76 degrees magnitude 33.82 N-s/m Fm =1.664 @ 14.46 degrees magnitude 1.664 N F=(Bl)E/Re =1.749 N equivalent source force with all electrical side referred to mech side. (Bl)^2/Re=8.89 N-s/m Electrical resistance referred to mechanical side.

In addition, as you requested:

F-((Bl)^2/Re)V =Fm =1.749-(8.89*0.492 @-71.76) =1.749-0.1369 +j0.415 =1.612+j 0.415 =1.665 @ 14.5 degrees (magnitude 1.665N) This agrees with Fm calculated using ZmV (as it should )

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

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

Bill W wrote:

D>-------------------

How did you arrive at 14.46 degrees here? TIA

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

D>> >The mechanical force is given by

Bill W. wrote:

D>------------

Don Kelly followed up re the above:

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

V Zm =Fm = 1.664 @14.46 N,

but

V Xmec = 1.664, not V Zmec

V Xmec = V wm-k/w = 0.0493*33.75 = 1.664

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

BLI = 7.17*.232 = 1.66344 (using your I of 0.232)

Matching your Fm of 1.664 within 0.03%. Not likely to be coincidence, but regardless I ran the figures on all 12 drivers equating

V Xmec = BLI V Xmec matches BLI on the 12 drivers within an average of 2.4%. Not bad, as many consider 3% matching to be excellent due to measurement tolerances.

Now your equation as equated to BLI is correct at resonance where current I and voltage E are in phase, but I do not think it should be applied at 227.4 Hz, since I is not in phase with E at 227.4 Hz.

Regardless, your equation would give zero actual or net force at resonance, and again we have no power Captain..

At resonance V Xmec = V wm-k/w = 0.157*zero = no force So unless I've missed something here, V*Xmec is not the expression for actual or net force, and the magnitude of net force at 227.4 Hz is not 1.664.

Bill W.

--------------------------- I calculated the velocity from BlE/Re =Zme V (Zme is the total impedance referred to the mech side. Then I looked at Fm=VZm where Zm is the mechanical impedance. and found the corresponding I Alternatively, I get the same result from E-RI =BlV =ZmeI where Zme is the mechanical impedance transferred to the electrical side. (or E=ZeI where Ze =Re +Zme) These calculations were given.The latter is

1.41 =(Re+(Bl)^2/[Rms+j(wm-K/w)]I =(5.88-j1.517)I giving I =0.232 @14.46 I showed the calculations and results. Shall I repeat them?

---------- I stand by what I said. Zm=2.32+j33.75 =33.82 @86.22 degrees V=0.0492 @-71.76 degrees The product F=1.664 @14.46 degrees The difference between 33.75 and 33.82 is 0.2% in magnitude. You also are using V=0.493 where I calculated a magnitude of V of

0.0492 -this is part of the difference. Do you really know V to an accuracy of 5 parts in 10,000 as implied?, or even 1 part in 1000? How about 3% which puts F somewhere between 1.61 and 1.71? You are approximating Zm by using Xmec and only calculating the magnitude of the force. Using your approach with |V|=0.0492 (my value calculated from the base data) I get 1.660 N magnitude. Since the phase angle is large, this approximation is good for the magnitude of the force. As I said before, the data doesn't support 4 figures so 1.66 is as good as you get for magnitude.

-----------

---------- Yes-it is the Lorentz force -- except that I have taken into account the phase. F=1.66 N. Saying 1.66344 is garbage.- you can't squeeze 6 figure accuracy out of 3 figure data even if your calculator says so.

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

------------ See above (and at the end) - VXmec gives a good approximation for magnitude at this and higher frequencies. What it does is use the approximation Zm =jXmec =Xmec @ 90 degrees. This will lead to about a 4 degree error in the phase of I

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

----------------- The whole purpose of my approach is because there are phase shifts and the approach takes these into account. At resonance it is not necessary simply because things are in phase. At any other frequency there are phase shifts and any approach used must account for these. Using the phasor methods as I have done is the normal method of AC circuit analysis as applicable in this case. Using magnitudes only is something that gets beaten out of students in their first AC circuits course.

My use of the equations is correct at any frequency for which the data is valid. If the initial data is correct, the results will be correct. In fact, what I have been trying to tell you is that, at 227.4Hz your approach approximates the magnitudes (and gives a good aproximation at this and higher frequencies) but doesn't account for phase effects. Working with magnitudes only causes problems. For example you had discrepancies in Eb as calculated from E-RI and from BlV because you worked only with magnitudes. - using the phasor methods gives consistent results as E-RI =Eb =BlV bloody well better be true whichever way you calculate it -if not it is a red flag that something is very wrong.

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

----------------- YOUR equation implies the above- this is in contradiction to your own data. VXmec is YOUR approximation, not mine, and it breaks down here. Please do not assign your errors to me as I am quite capable of making my own.

At resonance:

Zm =Rms+j0 at resonance so F=0.157(2.23)=0.35 N (at angle 0) {*****doesn't look like 0} Note that this agrees with the following Ze=Re+(Bl)^2/(Rms +j0) =5.78+23.05 =28..83 ohms (angle 0) (electrical input impedance) I=1.41/28.83=0.0489 A and corresponding F=7.17*0.0489 =0.351 N ***** Also Eb =1.41*23.05/28.83=1.127v (by voltage divider) so V=1.127*7.17=0.157 m/s **** (or Eb=1.41 -5.78*(0.0489+j0)=1.127V at angle 0)****

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

--------------- V*Xmec is not the actual force. V*Zm as a phasor quantity is. As for magnitude, I calculated a magnitude of 1.664 but recognise that 1.66 is as meaningful value as you will get. Zm has a magnitude of 33.82 Ns/m Xmec =33.75 N-s/m which is within 0.2% of Zm at this frequency. What you are effectively doing is saying Zm =0+j33.75 =33.75 @90 degrees as opposed to the true value of 2.23 +j33.75 =33.82 @86.22 degrees. It is a good approximation but using this approach at other frequencies leads to errors. In addition, with correct magnitudes and no phase information, further calculated results are nonsense. Using arithmetic based on magnitudes only can, and generally will, give incorrect results for AC conditions. The authors of your references recognise this but were trying to set up a simplified approach for a given purpose, recognising where the approximations are valid and also where they are not.

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

This gives mechanical power Pmec as equal to velocity multiplied by the terms other than velocity

Pmec = 0.0492 * 0.11 = 0.0054 watt or 5.4 mw I'll get to the 0.11 and 0.0054 magnitudes below.

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

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

Pmec = F cos angle v

where here: F = E Bl/Re = 1.41*7.17/5.78 = 1.75 cos angle = power factor = PF = 0.313 (as given by you), or PF = Rmec/Zmec = 11.12/35.5 = 0.313 (as given by me) v = 0.0493

therefore

Pmec = F PF v = 1.75*0.313*).0493 = 0.027

Rearranging Hallidays equation of Pmec = F PF v F*PF = Pmec/v = 0.027/0.0493 = 0.548

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

In other words

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

Now to relate net force to resistances

F net = (v Rms) + (v Rme) = (0.0493*2.23) + (0.0493*8.89) = 0.11 + 0.438 = 0.548

There is your 0.11 magnitude and as you can see, it represents only PART of the net or actual force.

where Rms = (2 pi fc m/Qmc = 2*pi*58.6*0.0253/4.17 = 2.23 = suspension mechanical resistance. Rme = (Bl)^2/Re = 7.17^2/5.78 = 8.89 = resistance encountered in overcoming back emf, which Small has called mechanical resistance of the motor, or sometimes called effective resistance of the motor.

Note phasor addition is not required as the powers are into resistance.

Now to net power:

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

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

Note this agrees EXACTLY with Halliday above.. Also note phasor addition is not required here as well, since this too is resistive.

I shall send the above for peer review.

This constitutes my entire reply in this post, leaving your derision to stand on its own.

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

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

----------------- So? Velocity has to be multiplied by something other than velocity to get power. Rm is N-s/m so Pm is given in (N-s/m)(m/s)^2 =Nm/s which is power The units are OK. I checked them before.

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

------------ No problem with Halliday but a problem with what you "repeatedly" (and incorrectly) said. In the problem as given, there is no other mechanical loss term. The Re term is not associated with mechanical loss. I have said this before. . So far we have only been considering a model of the driver (the model as indicated in Ludwig's site which gives the equivalent circuit. The only lossy elements are Re (electrical) and Rms (mechanical). Do you have data for other mechanical elements such as those shown by Ludwig (Lceb, Cmeb, Rel which are cabinet parameters)?

----------

-------------- Here there is a problem. I explained before how this force was arrived at. When you use F=BlE/Re you are taking the current source model (see Siskind or below) which can be represented by E/Re in parallel with Re (as I explained before). and moving this over to the mechanical equivalent to get F=BlE/Re driving the mechanical impedance Zm in parallel with (Bl)^2/Re The equation is then F=BlE/Re +{(Bl)^2/Re + Zm)V where Zm is the mechanical impedance. This becomes F=Fr +Fmec where Fr is the (Bl)^2(V/Re is due to the coil resistance as seen from the mechanical side. It is not an actual mechanical element although it is represented by one. (going the other direction, Rms converts to an electrical equivalent conductance (Bl)^2/Rms )

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

Note that your pf=0.313 agrees with the phase angle (71.76 degrees) of the total Zme = coil R moved to the mechanical equivalent ) +Zm where Zm =Rms

+j(Mw-K/w)

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

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

From this Pmec =2.23(0.0492)^2 =1.664*0.0492* (Pf) =5.4mW where Pf=2.23/33.75=cos 86.22 =0.066 Halliday's expression is fine- (it is a standard expression) provided you use the actual Zm.and actual mechanical force. It accounts for the phase. He probably used the approach that I have used but expressed it in terms of

pf. I have no argument with Halliday regarding this.

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

------

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

Ftotal =Fc +Fm where Fm is the actual mechanical force and Fc a force which is related to the electrical resistance Re. Note that Eb/Re =0.0611A which corresponds to a "force" (Fc) of

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

As to the current source model

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

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

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

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

------------------ Small is doing what I have done- simply expressed the electrical resistance in equivalent mechanical terms. I repeat, Rme doesn't represent an actual mechanical element. It represents the effect of coil resistance (NOT back emf related) and it must be separated from the mechanical resistance when calculating mechanical power. OK?

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

----------

---------- I agree with Halliday. You don't - as you have used a Z which includes the electrical element Re Note that the power in Re =|0.0216 =(Eb^2)/Re and is an electrical loss in the current source model. It is NOT, repeat, NOT, an actual mechanical power.

----------------------- No problem. . You are using |V|^2Rm correctly . Note however, that Halliday, when using pf is in fact using phasor methods. In other calculations, as I have shown, it is necessary to take into account phase relationships - the use of phasors is simply a very convenient way to do so. As you have shown by discrepancies, using magnitudes alone is not correct. I'm sure that Halliday realises this. I haven't read Halliday nor am likely to but I wonder how much explanation he gives for his formulae. Does he call Rme an actual mechanical resistance or does he call it the mechanical equivalent of the electrical resistance?

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

------- Better wait. There are too many inconsistencies in your work or interpretation of Halliday and others. If not, include my comments above.

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

------------- I'm was not deriding you. I think that I am not getting through to you. I have no reason to think that you are not good at what you do but I do think that your understanding of the theory behind the analysis is weak and too dependent on plugging in formulae without knowing the why of the equations and the approximations involved. Possibly Halliday and Small haven't made it clear as to the basis f their development. That is why I referred you to the Ludwig site as he looked at their work and showed the development of the model.

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

{*****doesn't

V=1.127*7.17=0.157

I simplifying your math into a magnitude to be apparent in my later equation. I have ask you to include magnitudes for better communication, but you continue to omit them. Why is that?

Sorry. I said I came here to learn, didn't I?

Ludwig's site which gives the equivalent circuit.

I am not going by Luswigs model. Never was. Rme = (Bl)^2/Re is included as a resistive (therefore lossy) term in models by Small, Beranek, Morse, Colloms, Olson, Keele, Villchur, Kinsler, Kloss, and Lahnakoski. It is used as an *equivilant* mechanical resistance, and their models are all essentially alike. Why do you use Ludwigs model?

Do all the math you like, but as I noted, it is applied mechanical force

You still don't acknowledge that 0.0054 is not the total net mechanical power???

This is not correct as I have shown. Again, do all the math you like, but here's the bottom line. Net mechanical power is just velocity squared times system *resistance*, analogous to electrical where P = I^2 R.

Pmec = v^2 * Rmec = v^2 * (Rme + Rms) = 0.0493^2 * (8.89 + 2.23) = 0.027 Perhaps this will help: Generally power is noted as force times velocity, if force and velocity are in phase, otherwise

Pmec = F v PF = 1.75 * 0.0493 * 0.313 = 0.027

Sorry, but there is no point in my addressing your following comments below until this is settled, as to who is correct on the magnitude of net mechanical power.

Others are welcome to comment on this.

Bill W.

--------------- Excuse me? I did post the magnitudes that you asked for. In addition, every time that I used the phasor form, I expressed it in both rectangular and polar forms. The polar form, as I explained, is (magnitude) @ angle so the magnitude is there loud and clear. Anyhow, I repeat what I sent (and it was sent) below.

************

Actually the magnitudes of the terms are given in my original calculation. I showed the polar form which is (magnitude) @ (angle) as well as rectangular form (real) +j(reactive) For data I used only the values of E=1.41 volts, Re=5.78 ohms, Bl=7.17 Newtons/Ampere or volt-sec/meter Rms=2.23 N-s/m, M=0.0253Kg and K=3425 N/m as I had indicated.

The following were calculated . V=0.0492 @ -71.76 degrees magnitude 0.0492 m/s Zm =33.82 @ 86.22degrees magnitude 33.82 N-s/m (NB I did have the angle wrong before) Fm =1.664 @ 14.46 degrees magnitude 1.664 N F=(Bl)E/Re =1.749 N equivalent source force with all electrical side referred to mech side. (Bl)^2/Re=8.89 N-s/m Electrical resistance referred to mechanical side.

In addition, as you requested:

F-((Bl)^2/Re)V =Fm =1.749-(8.89*0.492 @-71.76) =1.749-0.1369 +j0.415 =1.612+j 0.415 =1.665 @ 14.5 degrees magnitude 1.665N

This agrees with Fm calculated using ZmV (as it should )

************ Going a bit further, I =Fm/Bl =0.232 @ 14.46 degrees magnitude 0.232A Eb =(Bl)V =0.353 @ -71.76 degrees Magnitude 0.353 volts check: E-RI=1.41 -(5.78*0.232) @ 14.46 degrees =0.353 @71.6 degrees OK

The results are consistent and do satisfy the original basic equations Note I is not 1.75/7.17 =0.244 magnitude (which is E/Re independent of frequency and mechanical load) -

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

--------- I realise that - but he has put it in a nutshell (and has based his work on Small etc)

----------

---------- I don't have the other's and Ludwig's model, based on Small's work presents things in a way that is, to me, fairly intuitive. From what you have quoted, it appears that Small and others actually have used a similar model to develop their final simplified models. I have no problem with it being an "equivalent mechanical resistance" and have said as much.

"This becomes F=Fr +Fmec where Fr is the (Bl)^2(V/Re is

However, it is a reflection of the electrical resistance and is not part of the actual mechanical impedance which is due to only the mechanical elements. It is useful in analysis.>

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

-------------- Bull. It is a electrical current source equivalent expressed in mechanical terms. Does representation of the mechanical parameters in terms of equivalent electrical parameters make them actual electrical elements and with losses that are no longer mechanical but electrical? I could bring Rms over to the electrical side (it will be 23 ohms ) and then claim that there are no mechanical losses -just electrical losses. Do you see the flaw in this? Just because the mechanical resistance can be expressed in terms of an electrical resistance, does it follow that the losses are then actually electrical losses and not mechanical losses at all?. That is illogical . Yet your argument shows the same illogic.

Also note that from the force you have given, I =1.75/7.17 =0.244 and this is the same as E/Re =1.41/5.78 =0.244 A Is this the actual current ? Also this current and the corresponding force are independent of the mechanical impedance and the frequency. Does that make sense to you? Do you expect the force and the current associated to be constant? Is the current at resonance the same as the current at 227Hz? If not then there must be something wrong with the assumption that the actual mechanical force is

1.75N. Alternatively the "force" of 1.75 N could possibly be other than the actual mechanical force applied to the mechanical system. That is my position based on straight- forward circuit analysis (incidently no phasors needed for this ) .

Back to basics

We have E-RI =Eb for the electrical system where Eb is the back emf =(BL)V We also have Fmec =ZmecV where Zmec depends on the various masses, spring compliances and damping factors of the mechanical system. and V is the mechanical velocity This expression deals only with the actual mechanical dynamics of a force acting on a mass, spring,damper system and does not deal with any electrical factors per se. .

(In this case Zmec =Rms +j(wM-K/w) = sqrt(Rms^2 +(wM-K/w)^2) @ angle arctan (wM-K/w)/Rms )

These are the basic equations which you seem to have accepted If you haven't then it is back to square one.

Rewrite the electrical equation (E-Eb)/Re =I In addition Fmec =(Bl)I

so Fmec =(Bl)I =(Bl)E/Re -(Bl)Eb/Re =ZmecV

but (Bl)Eb/Re = (BL)^2V/Re

so Fmec ={(Bl)E/Re -(Bl)^2V/Re} = ZmecV ***

or (Bl)E/Re =[(Bl)^2/Re +Zmec]V

(Bl)E/Re is not the actual mechanical force (Fmec ) but it as well as the (Bl)^2/Re term make up a non-ideal current source referred to the mechanical side. This approach is convenient for determining V and from V , Fmec can be found. From V and Fmec, the actual mechanical power can be found as well as the electrical current and the back emf.

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

n't acknowledge that 0.0054 is not the

------------ I certainly don't and I have given my reasoning for that. You have not given reasoning for your contention. You have made quotes but the one you have given re (BL)^2/Re supports my contention- not yours. I have stated that it is the electrical resistance term referred to the mechanical side - as such, even though it is an "equivalent mechanical resistance" it is not an actual mechanical resistance. There is a very important difference. -----------------

--------------------- First of all, you haven't shown anything. You have made assertions without providing the reasoning for your stand. Sorry but that doesn't cut it. When you quote references, your quotes support my position as well as, or better, than they support your position. Since the basis of any model of an electromechanical transducer is, of necessity, for working purposes, mathematical it is hard to avoid using mathematical methods- hand waving doesn't work. (Small, et al didn't avoid mathematical methods in their developments). I have shown a different approach to the basic equations ( above) . Go through it step by step, not worrying about phasors. It is not difficult math (you have shown that you can handle more difficult stuff) but it does show where the terms in contention comes from. Then come back with questions about it.

Noting that the term (Bl)^2/Re depends on an electrical resistance, how can it involve actual mechanical power? This term along with (Bl)E/Re make up the equivalent current source "as seen from the mechanical side". Since (Bl)^2/Re is simply a consequence of the model and the electrical resitance and is not an actual mechanical element , the power into it , while expressed in mechanical terms is not a an actual mechanical loss but a loss in the electrical side equivalent to Eb^2/Re. Unfortunately this does not translate into I^Re for reasons which I won't get into.

----------- no kidding. I have nothing against the form of that equation but I do have a problem with the values used. The power that you have given is the mechanical power plus an internal loss in the current or force source.

----------

-------------- Fair enough -but at least read what I have said and the basis for the inclusion of the (Bl)^2/Re term as well as the difference between the (Bl)E/Re "force" and the actual mechanical force. Note that none of the "math" that I have used is anything special - It is simply application of circuit analysis to an electromechanical system - quite commonly done. It is also obviously what Small etc have also done. I referred to Ludwig because it was the only source that I could find on the net that actually had some meat and went back to get show the basis of Small's work. I would love to see Small's original paper. The others seem to involve unspecified programs that would provide a model when the data was fed in - essentially someone has taken the math and used it to write a program. Other sites give the way to make certain measurements and provide a program to caculate the parameters.

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

(Bl)^2/Re=8.89 N-s/m Electrical resistance referred to

Thank you, however the above still is not explicit at all without your derivations and individual magnitudes. For example, I would have done the notation for velocity so: _ _ v = F/Zmec = 1.75/35.5 = 0.0493

i.e. I need the derivation and all magnitudes. Now if you did this earlier as we went along, I should have noted such, and sincerely apologize. Therefore would you please fill in the derivation and individual magnitudes for *all* your parameters above? TIA

Note, that I have generally provided such detail in each ongoing equation of mine.

Bull back. The magnitude is real and correct, and my equation is quite intuitive. Kloss and Small have stated that the equation gives mechanical force generated, i.e. force applied. I believe you can't see the trees for the math, or something like that... I have given my view on this before, here it is again.

Applied force means just that, it means the gross (not net) force applied without considering any load reaction due to motion. It is *not* BLI as used during motion, it is BLI with a blocked coil (no motion), and written as _ _ F applied = BL E/Re = 7.17*1.41/5.78 = 1.75.

Do you see this? I visualize this as "freezing time" at the instant of applying voltage (before motion can occur - time increment of ~ .0000000137 ms) +/- 1.75 dB)... :-) At this point we are dealing with the DC resistance of the coil Re = 5.78 ohms, voltage applied E = 1.41 volts, and the motor force factor BL = 7.17, so force *being applied* again is _ _ F applied = BL E/Re = 7.17*1.41/5.78 = 1.75.

Again, can you see this? If not, we'll likely never get past square one.

used the approach that I have used but expressed it in terms >of

given re (BL)^2/Re supports my contention- not yours. I have stated that it

Before I answer this one, what is your derivation for (BL)^2/Re = 8.89 ? TIA

Then I shall give my derivation and reasoning and we'll see how they fit in, and how intuitive they are.

I'm beginning to think most anything I offer will be mathed into disapproval. :-) Let me offer the view on power from the topmost authorities, perhaps you'll believe them, and abandon LooseWigs theory... .. er Ludwig's that is.. Sorry - bad joke.. :-(

OK here we go.....

Halliday and Morse:

P = F * v * cos angle = 1.75*0.0493*0.313 = 0.027

Beranek and Villchur:

P = v^2 * Rmec = 0.0493^2*11.12 = 0.027

Colloms:

P = F^2/Zmec^2 * Rmec = 1.75^2/35.5^2*11.12 = 0.027

Kinsler:

P = F^2/Zmec cos angle = 1.75^2/35.5*0.313 = 0.027

Note that all agree **exactly**, giving magnitude for net mechanical power of 0.027, not 0.0054 as you claim.

Now for the clincher: Using my *measured* amplitude (equilibrium to end point of motion) at 227.4 Hz of 0.0000543, and the work equation P = work/time, then for 1/4 cycle

P = F*d cos angle/t = 1.75*0.0000543*0.313/0.0011 = 0.027

Note d is *measured*, time for a quarter cycle is a given at T/4 = 1/f divided by 4, cos angle was provided by you, then *applied* force *has* to be 1.75.

Can you see this now? Square one forever? :-)

Bill W.

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

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

former posting:

(Bl)^2/Re=8.89 N-s/m Electrical resistance referred to

Thank you, however the above still is not explicit at all without your derivations and individual magnitudes. For example, I would have done the notation for velocity so: _ _ v = F/Zmec = 1.75/35.5 = 0.0493

i.e. I need the derivation and all magnitudes. Now if you did this earlier as we went along, I should have noted such, and sincerely apologize. Therefore would you please fill in the derivation and individual magnitudes for *all* your parameters above? TIA

Note, that I have generally provided such detail in each ongoing equation of mine.

cal force (Fmec ) but it as well as the

Bull back. The magnitude is real and correct, and my equation is quite intuitive. Kloss and Small have stated that the equation gives mechanical force generated, i.e. force applied. I believe you can't see the trees for the math, or something like that... I have given my view on this before, here it is again.

Applied force means just that, it means the gross (not net) force applied without considering any load reaction due to motion. It is *not* BLI as used during motion, it is BLI with a blocked coil (no motion), and written as _ _ F applied = BL E/Re = 7.17*1.41/5.78 = 1.75.

Do you see this? I visualize this as "freezing time" at the instant of applying voltage (before motion can occur - time increment of ~ .0000000137 ms) +/- 1.75 dB)... :-) At this point we are dealing with the DC resistance of the coil Re = 5.78 ohms, voltage applied E = 1.41 volts, and the motor force factor BL = 7.17, so force *being applied* again is _ _ F applied = BL E/Re = 7.17*1.41/5.78 = 1.75.

Again, can you see this? If not, we'll likely never get past square one.

given re (BL)^2/Re supports my contention- not yours. I have stated that it

Before I answer this one, what is your derivation for (BL)^2/Re = 8.89 ? TIA

Then I shall give my derivation and reasoning and we'll see how they fit in, and how intuitive thay are.

I'm beginning to think most anything I offer will be mathed into disapproval. :-) Let me offer the view on power from the topmost authorities, perhaps you'll believe them, and abandon LooseWigs theory... .. er Ludwig's that is.. Sorry - bad joke.. :-(

OK here we go.....

Halliday and Morse:

P = F * v * cos angle = 1.75*0.0493*0.313 = 0.027

Beranek and Villchur:

P = v^2 * Rmec = 0.0493^2*11.12 = 0.027

Colloms:

P = F^2/Zmec^2 * Rmec = 1.75^2/35.5^2*11.12 = 0.027

Kinsler:

P = F^2/Zmec cos angle = 1.75^2/35.5*0.313 = 0.027

Note that all agree **exactly**, giving magnitude for net mechanical power of 0.027, not 0.0054 as you claim.

Now for the clincher: Using my *measured* amplitude (equilibrium to end point of motion) at 227.4 Hz of 0.0000543, and the work equation P = work/time, then for 1/4 cycle

P = F*d cos angle/t = 1.75*0.0000543*0.313/0.0011 = 0.027

Note d is *measured*, time for a quarter cycle is a given at T/4 = 1/f divided by 4, cos angle was provided by you, then *applied* force *has* to be 1.75.

Can you see this now? Square one forever? :-)

Bill W.

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

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

I had to clip off the former posting, supernews editor full or somesuch. I saved it in case you want to refer to it.

Bill W.

(Bl)^2/Re=8.89 N-s/m Electrical resistance referred to

Thank you, however the above still is not explicit at all without your derivations and individual magnitudes. For example, I would have done the notation for velocity so: _ _ v = F/Zmec = 1.75/35.5 = 0.0493

i.e. I need the derivation and all magnitudes. Now if you did this earlier as we went along, I should have noted such, and sincerely apologize. Therefore would you please fill in the derivation and individual magnitudes for *all* your parameters above? TIA

Note, that I have generally provided such detail in each ongoing equation of mine.

the actual mechanical power can be found as well as

Bull back. The magnitude is real and correct, and my equation is quite intuitive. Kloss and Small have stated that the equation gives mechanical force generated, i.e. force applied. I believe you can't see the trees for the math, or something like that... I have given my view on this before, here it is again.

Applied force means just that, it means the gross (not net) force applied without considering any load reaction due to motion. It is *not* BLI as used during motion, it is BLI with a blocked coil (no motion), and written as _ _ F applied = BL E/Re = 7.17*1.41/5.78 = 1.75.

Do you see this? I visualize this as "freezing time" at the instant of applying voltage (before motion can occur - time increment of ~ .0000000137 ms) +/- 1.75 dB)... :-) At this point we are dealing with the DC resistance of the coil Re = 5.78 ohms, voltage applied E = 1.41 volts, and the motor force factor BL = 7.17, so force *being applied* again is _ _ F applied = BL E/Re = 7.17*1.41/5.78 = 1.75.

Again, can you see this? If not, we'll likely never get past square one.

given re (BL)^2/Re supports my contention- not yours. I have stated that it

Before I answer this one, what is your derivation for (BL)^2/Re = 8.89 ? TIA

Then I shall give my derivation and reasoning and we'll see how they fit in, and how intuitive thay are.

I'm beginning to think most anything I offer will be mathed into disapproval. :-) Let me offer the view on power from the topmost authorities, perhaps you'll believe them, and abandon LooseWigs theory... .. er Ludwig's that is.. Sorry - bad joke.. :-(

OK here we go.....

Halliday and Morse:

P = F * v * cos angle = 1.75*0.0493*0.313 = 0.027

Beranek and Villchur:

P = v^2 * Rmec = 0.0493^2*11.12 = 0.027

Colloms:

P = F^2/Zmec^2 * Rmec = 1.75^2/35.5^2*11.12 = 0.027

Kinsler:

P = F^2/Zmec cos angle = 1.75^2/35.5*0.313 = 0.027

Note that all agree **exactly**, giving magnitude for net mechanical power of 0.027, not 0.0054 as you claim.

Now for the clincher: Using my *measured* amplitude (equilibrium to end point of motion) at 227.4 Hz of 0.0000543, and the work equation P = work/time, then for 1/4 cycle

P = F*d cos angle/t = 1.75*0.0000543*0.313/0.0011 = 0.027

Note d is *measured*, time for a quarter cycle is a given at T/4 = 1/f divided by 4, cos angle was provided by you, then *applied* force *has* to be 1.75.

Can you see this now? Square one forever? :-)

Bill W.

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

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

Problem... supernews, editor full or somesuch. Clipped former posting below here- saved to a file if you wish to refer to it.

Bill W.