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]> 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. > > > -- > > > -- > -- *Forrest Christian* *CEO**, PacketFlux Technologies, Inc.* Tel: 406-449-3345 | Address: 3577 Countryside Road, Helena, MT 59602 [email protected] | http://www.packetflux.com <http://www.linkedin.com/in/fwchristian> <http://facebook.com/packetflux> <http://twitter.com/@packetflux>
