1998 Grand Caravan

Page 10 of 13  
Budd Cochran wrote:


If only you knew how to read it. I'm quite sure it has Ohm's law spelled out correctly.
Matt
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And yet you claim to not denigrate a person's education.
You are a lying hypocrite!
I've had enough of you.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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Budd Cochran wrote:

I didn't denigrate your education. I denigrated your ability to read. You posted a blatantly incorrect physics relationship, yet you claim to have a book that in all likelihood would have the correct rendition of Ohm's law. So, this suggests that you either can't read or can't understand the book you have.

Only in your feeble mind.

It is about time. I'm surprised you could tolerate truth and correctness even this long, given your propensity for fiction.
Matt
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arrogant
And then there are those that can never admit to error, even when every fact screams that you are.

The load test proves them right and you wrong.

What makes you think that engineering had anything to do with it. How about accounting.

Terrific,
design
It does no such thing. The fuse is there to prevent a fire, not eliminate a voltage drop.

fuse
W R O N G ! ! !

W R O N G ! ! !

The funny thing is that the engineers probably had nothing to do with it. Many of the failures in today's automobiles are due to cost cutting, not engineering.

of
the
W R O N G ! ! !

W R O N G ! ! !

As usual, you don't know WTF you are talking about and only see 1/1000th of the picture.
--
If at first you don't succeed, you're not cut out for skydiving



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What you are failing to grasp here is that a load test was done, even though all of the engineers said it was useless. Your statement above is exactly why I said a load test was needed; it is definitive proof of where the problem lies. If you go back and read what I've said on that thread, I didn't claim to know exactly where the problem was, only that a load test was essential in finding it.
They claims, as well as yours, to the contrary are proven wrong by the fact that a load test was in fact done. The claim that 629A (or whatever amperage was claimed) was found means nothing until a manufacturers CCA rating for the battery is posted, and I haven't seen anyone post that as a "fact" yet.
The fact that the wiring is determined to be the center of this problem means that one group of engineers is wrong, either its you internet experts, or the ones in the Chrysler engineering and design studios. As I said before, and especially in light of your outrageous claims (helium has no weight?) I'll go with the Chrysler guys as more reliable.
--
Max

"There are four boxes to be used in defense of liberty:
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Budd Cochran wrote:

(But yet it has plenty of power to start the car - an order of magnitude larger load - oh - I'm sorry - that's secret engineering talk for at least ten times as much)

OK - so you both are totally ignorant of resistive drops and the real world effects of Ohm's law. Line drop is resistive drop over wire. With increase in current, it's effects in voltage change are, for all practical purposes, instantaneous (on the order of nanoseconds or even quicker - throw in some parasitic inductance and capacitance, and you might get into the high microsecond range - so like I said - for our discussion, it's instantaneous).
So I'm left with the dilemma of figuring out if you guys are truly this dumb or if it's an act just to see how long you can keep this stupid conversation going (you did talk about "playing" people to get your jollies earlier, so maybe that's it).
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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Naw, I ain't got no axperance a-tol. I jes werked on 36 and 48 volt lift truck control and power circuits fer ten yars.
You know, the kind where a current of a couple amps regulates the speed of a motor requiring half a thousand amps at full speed and 4 thousand amps at stall. The kind with individual battery cells weighing 200 lbs each.
The kind where you can plug in a set of welding cables and burn 1" steel in a single pass. Ya wanna talk orders of magnitude? How's that for magnitude, Bill?
It is you that is confusing primary circuit current with high amp starting current. And failing to mention that as current ( amps) rises, voltage loss decreases in a circuit. And I believe you're doing it deliberately to save face.

Well, I've come to the decision your diplomas aren't in automotive fields or anything resembling them. Which makes you no more an authority than Parker was with his sociology PhD.
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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Budd Cochran wrote:

More of that "new" physics, eh? I learned that the voltage drop across a resistance was related to current by V=IR, commonly called Ohm's law. A circuit is a resistance, maybe a small one if the wire is large enough, but a resistance nonetheless. So as current increases, with resistance remaining the same, the voltage drop around the circuit will increase, not decrease.
Can you explain how you have found a way around Ohm's law?
Matt
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Matt Whiting wrote:

To paraphrase the banditos in Blazing Saddles: "Ohm's Law!! We don't need no stinking Ohm's Law!!"
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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Matt Whiting wrote:

Hi...
Perhaps its been repealed? :) :)
Not to fuel the fire, but in a (perhaps futile) attempt to make you fellows come to some kind of a consensus...
I have a 1500 watt electric heater in my garage, for when I work a bit on my Chrysler during a Winnipeg winter...
Everytime I plug this heater in, it gets nice and toasty warm, which is good. But - the durned wire gets very, very warm, and concers me a bit. I wonder how, and why, and where that bit of energy comes from.
Again, not to fan the flames, but just food for thought.
Take care.
Ken
PS. I coulda spelt ohm backwards and really messed the conversation up :)
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Ken Weitzel wrote:

Are you being sarcastic or do you really now know what causes the heating in the wires in your heater? Or do you mean the cord to the heater? Either way, it is the same thing. Power = IsquaredR. The cord to the heater has some resistance and it is carrying a lot of current for a 1500 W heater, so it will get warm and dissipate heat just as the heating element wires in the heater get hot and dissipate heat.
Matt
Matt
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Matt Whiting wrote:

Well - sure - if you believe Ohm's law!!! What poppycock!
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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Ken Weitzel wrote:

That would be like asking me to come to a concensus that the sun rises in the west and sets in the east. Ain't gonna happen.

We're back to resistance in the wire which - according to Ohm's law - causes a voltage drop proportional to current flow.
Now - power loss (heat generated in that wire) = current^2 * resistance. It also is voltage^2 resitance. Either way you calculate it, it comes out the same. (For Budd and Max, the '^' is a mathemeatical/engineering symbol - brought about by the use of computers because superscripts are awkward in computers - meaning 'to the power of', or in this case. '^2' means 'squared'. So you'll understand: Power lost is equal to voltage squared divided by resistance, where power is in watts, voltage is volts, and current is amps, or some other compatible units.)
[sarcasm]Now - if you do not believe Ohm's law, but instead believe Budd's law (Budd's law: "as current ( amps) rises, voltage loss decreases in a circuit" - a direct quote from the man himself), the resistance in the wire causes a voltage *rise* with current - so if the voltage at the wall outlet is 115, the voltage reaching the heater is actually higher - say 120 volts. So, since the current is the same in the series circuit, instead of the resistance causing power loss, power is actually generated by the resistance in the wire (you just think the wire is heating up, Ken - actually, by Budd's law, the wire is getting colder. If you are smart, Ken, you will buy a whole bunch of heaters and tap into this power generation and sell off the excess power.[/sarcasm]

No - with a simple real world example, you've provided another opportunity to illustrate gross and willful ignorance on the part of certain people.

Believe it or not, there is an engineering unit called "mhos" (pronounced like the name of the leader of the Three Stooges) which is the inverse of resistance (i.e., 1/ohms) - it is called conductivity. So if a certain wire or junction or device is a good conductor, it will have high mhos.(This last paragraph is not tongue-in-cheek.)
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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(Sarcasm mode at full throttle, full boost and full NOX)
Oh, I keep forgetting, only an "engineer" like yourself ( ROFLMBO!!!) can deliberately confuse a topic.
(sarcasm mode disengaged)
--
Budd Cochran

John 3:16-17, Ephesians 2:8-9
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Budd Cochran wrote:

Ha, ha, ha... Ohm's law is about as simple as it gets. Only you could be confused about E=IR.
Matt
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Matt Whiting wrote:

Hi Matt...
Wonder if someone could if they didn't know the abbreviations... so just in case...
E = voltage (in volts) I = current (in amps) R = resistance (in ohms)
Take care.
Ken
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Is this your coded way of saying you were wrong.
--
If at first you don't succeed, you're not cut out for skydiving



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Budd Cochran wrote:

OK - so I was right - you are in fact totally ignorant of Ohm's law (that's the thing that says that as current rises, voltage drop *increases*, system voltage decreases - you got it exactly backwards). Take your above statement ("as current ( amps) rises, voltage loss decreases in a circuit") and show it to one of your technically competent buddies, maybe one of the few engineers for whom you have respect, and see what he says. If he truly is competent, he will tell you that you are making a fool of yourself. Oh - and have him explain Ohm's law to you - have him show you what happens to voltage drop across a resistor as current increases.
Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')
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practical
into
And your spelling shows just how much you really know.

a
What in the hell does this shit have to do with anything?

in
magnitude,
Oh, gee Budd, you're so cool, can we hang out with you? What exactly does this have to do with the discussion?

What exactly does this primary - secondary bullshit have to do with a weak battery. Oh yea, not a damn thing. If the battery is so weak as to drop whole volts in a few seconds with a 10 amp load, how could it possibly have enough power to start the engine when the load on it is over 100 amps, especially when the computer that controls that engine also needs better than 10.5V to fire the injectors and this has nothing to do with a resistive voltage drop at all.

Hahahahahahaha, could you be any more wrong? I guess that Ohms law has no meaning in your world. You do know the simple formula E = I x R, right????? That same formula is used to calculate voltage drop as well. To calculate voltage drop, R is the resistance of the connections and wires and I is the current flowing thru the circuit. Now please explain how if R is held steady and I increases, how E can go down given the above formula. Don't worry, I'll wait.

or
The sad thing is that they are correct while you and Max are just making fools out of yourselves.
--
If at first you don't succeed, you're not cut out for skydiving



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though
LOL, sorry Maxi but you are still wrong. A simple voltage test on the battery while the inverter was plugged in was all that was needed. You really do flip flop more than Kerry. First it was a bad battery, then it wasn't but the load test was necessarry, then back to the battery was bad again. Make up your mind because either way, you are still wrong. The battery was not bad and a load test was not necessary.

fact
amperage
yet.
I must say that it funny to watch you twist and squirm rather than just admit to your error or just shut the hell up. Since it is a mini-van with a factory battery, do you really think it is going to be anything over 700 CCA and I see you are still flip-flopping. In the first paragraph you claim that the load test confirming the battery to be ok proves you right and now you have managed to flip flop back to the battery possibly being bad again. If you actually new what a voltage drop was you could possibly have an argument that makes sense.

experts,
Once again, you can't see the forest thru the trees. What makes you think that the engineers caused this problem? Perhaps it was the cost cutters or perhaps they were not expecting such a device to be connected there or saving money took a priority like in the Dakota front ball joints or using the NV3500 in the full sized RAM even though it reduced towing capability or using undersized steering shaft joints that tend to fail prematurely or putting the damn fuel filter in the tank as part of the pump. Yea Max, cost savings is never placed over full reliability or longevity.
--
If at first you don't succeed, you're not cut out for skydiving



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