Looking for Forums for Electronics Engineering Design, etc.

What other Forums or Newsgroups are on the net for engineers who specalize in...
- Electronic Circuit Design - System Design
- Antenna Design
Thanks for any info.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
----- Original Message -----
Newsgroups: alt.engineering.electrical Sent: Thursday, December 11, 2003 11:15 AM Subject: Re: Motor torque and back emf

remove the urine to answer
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

------ shipping ---------
Bill W wrote:

Don Kelly wrote:

How did you arrive at 14.46 degrees here? TIA
--------------------
Don Kelly wrote:

Bill W. wrote:

Don Kelly wrote:

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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
remove the urine to answer
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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) - -------------------------------

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.