Re: Questioning the math Re: reluctance
- From: Spehro Pefhany <speffSNIP@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 25 Oct 2006 11:37:11 -0400
On Wed, 25 Oct 2006 10:00:10 -0500, the renowned "amdx"
<amdx@xxxxxxxxxxx> wrote:
The URL below adds iron to an air core solenoid to reduce
the reluctance and increase force.
The author calculates the reluctance of an air core as R1.
He then calculates the reluctance of two iron end plate washers as R2 and
R4.
Then the iron tube around the air core as R3.
He then adds R1+R2+R3+R4 for total reluctance.
It seems to me that R1 (the air core) is eliminated because it is
replaced (shunted) by the iron.
Can someone explain what I am missing here.
I understand at some point there will be an air gap,
but it will much smaller, I'll deal with that later.
http://www.oz.net/~coilgun/theory/externaliron.htm
Also in the section, "R2,R4: Finding Reluctance of a Flat Washer"
He goes through some calculus and gets,
Reluctance = 1/(2*pi*t*u)[ln(r2)-ln(r1)]
You need another pair of brackets to make the above correct:
Reluctance = (1/(2*pi*t*u))*[ln(r2)-ln(r1)]
= [ln(r2)-ln(r1)]/(2*pi*t*u)
He then reduces this to,
Reluctance = (ln(r2/r1)) / (2*pi*t*u)
I don't see this as correct,
If you recall your math, ln(A) - ln(B) = ln(A/B) right?
Just as e^a * e^b = e^(a+b)
(related to how slide rules work...)
Can I get a little help?
Thanks,
Mike
.
Best regards,
Spehro Pefhany
--
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