I know right, whats up with that Can?
On 3/10/2017 8:51 PM, Forrest Christian (List Account) wrote:
I cheated and used microsoft mathematics (free) to simplify. Could
have done it by hand, but when you have a decent computer algebra
system which doesn't break rules, why not use it?
(V+-sqrt(v^2-4*r*p))/2 is effectively what I came up with. not sure
about the 2*-R in your solution as that seems odd...
The v-sqrt(..... solution gives nonsensical answers so the
(V+sqrt(v*v-4*r*p))/2 is the final formula I came up with.
I sort of had a head start as I already worked through this a while
back with some DC power site coursework I was working on. That and
some web-based tools for DC power systems. I would have just grabbed
it and threw it out here but I'm already in memphis...
In other news, I still don't get Elvis, and Memphis tamales taste
exactly like those in a Nalley or Hormel tamale can.
On Fri, Mar 10, 2017 at 8:43 PM, Chuck McCown <[email protected]
<mailto:[email protected]>> wrote:
Also, I think Forrest was working on a solution that would not
require a quadratic solver. I did not study his post closely, but
there should be a way to solve the lower root with the quadratic
equation.
-V +/-[ sqrt (v^2-4*-r * –P)]/2*-R
Is this similar to what you did Forrest?
*From:* Chuck McCown
*Sent:* Friday, March 10, 2017 7:39 PM
*To:* [email protected]
*Subject:* Re: [AFMUG] Fw: the solution
OK, so since it is proven, again here it is:
0= – RI^2 + VI – P
With the coefficients being:
A= –R (wire loop circuit resistance)
B= V (power supply voltage)
C= –P (power of the constant power load)
Enter those in a quadratic solver and if it does not give complex
roots it will work.
The current will be the lower root.
Now if someone smarter than me will please figure out how to solve
for the minimum voltage that you can use. I feel in my bones it
is the first derivative but I cannot understand how that could be
true because the power coefficient drops out. So obviously that
is not the path to the solution. However when you close in on the
minimum workable voltage the quadratic roots converge to the same
number.
Perhaps it is a limit type of problem.
*From:* David Milholen
*Sent:* Friday, March 10, 2017 7:31 PM
*To:* [email protected]
*Subject:* Re: [AFMUG] Fw: the solution
Ahh yes.. AKA the LOOP :)
On 3/10/2017 9:07 AM, Chuck McCown wrote:
Because of the feedback.�
�
*From:* David Milholen
*Sent:* Friday, March 10, 2017 6:02 AM
*To:* [email protected]
*Subject:* Re: [AFMUG] Fw: the solution
�
I love this stuff I feel it still keeps me sharp in my later years :)
My question is why is it NOT Linear? Would it be because of the
Variables in the load and wire?
�
On 3/9/2017 10:03 PM, Chuck McCown wrote:
-----Original Message----- From: Chuck McCown Sent: Thursday,
March 09, 2017 9:00 PM To: [email protected] Subject: Re: [AFMUG] Fw:
the solution
This was a PDF.� I can't remember if PDF come through the list
or not.�
-----Original Message----- From: Chuck McCown Sent: Thursday,
March 09, 2017 8:59 PM To: [email protected] Subject: [AFMUG] Fw: the
solution
As you can see, I actually arrived at the solution early on, but
then stumbled around searching for the linear solution which
does not exist.
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