1998 Grand Caravan

How much did I win???

Agreed. I've mentioned before about having worked with a few Big Four (at the time, GM, Ford, Chysler, AMC) on developing the cat converters back in the early 70's. I was impressed that not a one of them was like Matt or Bill.

You noticed that too? ;) The bad part? One has an Electrical degree.

It's curious also that Mopars have never been known for electrical problems to the extent of Ford or GM.

The simplest description of a fuse is "a load sensing automatic, non-auto resetting switch". It blows, it stops current into everything downstream.

And the undervoltage limiter shuts it all down.

Agreed.

There is the third variable . . .the owner installed an oversized fuse and/ or a factory assembler screw up and installed an oversize fuse. Quick check: read the fuse rating and then see what the fuse is rated at.

Max, as we both know, too often idiocy or ego is considered to be more important than common sense or applied education.

My friend, there is no doubt that Chrysler engineering has its act together, but our resident engineers are just poor actors.

(you later go with the second choice, therefore...) As has been pointed out several times, if you are correct, the battery can't maintain voltage with a 10 amp load, but it can start the car without any problem wich requires several times the 10 amp load. Not plausible.

Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')

(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')

An as I've stated numerous times, the inverter shut down due to voltage drop. The starter motor is too "dumb" to care; it will continue to crank at less than 9volts, not very well, but it WILL crank. As such, if the voltage dropped to 11v nominal, and held long enough to spin the engine for one firing stroke, and the ECM fired one injector, the engine is starting on 11v start and running on 13.5+ after alternator reaches idle speed. We've all heard this happen in our youth while in the cold and with a near dead battery, while uttering words similar to "f***, my ____ is gonna kill me if I can't get home." This could easily happen in the time mentioned as that which it took the inverter to shut down. As was noted, he had better than

11v at the port under this load. Since there IS some drop in the line, its almost sure that he had more at the battery, but not by much.

As such, your engineering which says it cannot happen (and I'd agree on first glance) is overridden by reality where imperfect voltage doesn't necessarily mean things come to a dead stop. Remember, there is NO switch on the starter that is operated by a voltage sensor. The only thing switched is the injector, needing 11v nominal to fire.

IOW, the engine needs 11v to start, the inverter needs 12v to operate.

And what are your degrees in?

What gauge wire carries the starting current? The power outlet current? The pilot current for engaging the solenoid is a mere fraction of the current required to crank the engine . . .your argument is false.

And your ego is writing checks of a magnitude above what your education can pay.

Primary (control circuit) current for a starter is more than a magnitude less than the cranking current.

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.

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

(1) I already answered that. (2) You already stated that you wouldn't believe me no matter what I told you.

Except (even though you don't realize it), what you've been arguing is that the drop is coming from within the battery (internal resistance) - and that drop will be due to *total* current it is supplying at any given moment (again you illustrate your gross ignorance of reality as dictated by Ohm's law). So, when engaging and cranking the starter, the drop in the battery will primarily be determined by the current that the starter is sucking. Even with zero drop across the solenoid supply wire, the voltage it sees will be no higher than what the battery can provide at the heavy current draw of the starter.

Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')

I thought the cutoff voltage of the inverter was 10.5.

Bill Putney (To reply by e-mail, replace the last letter of the alphabet in my address with the letter 'x')

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')

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')

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 :)

And your spelling shows just how much you really know.

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

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.

The sad thing is that they are correct while you and Max are just making fools out of yourselves.

LOL, you really are too much Max. Many times failure to start has been directly attributed to the battery and if a simple 10A load will pull the battery down to a point where the inverter shuts down, there would be no way in hell there would be enough voltage under a 150+ amp draw to fire any injector, never mid starting it time and time again and as you seem to have forgotten once again, the battery WAS LOAD TESTED and is not the problem, no matter how desperatly you may want it to be.

Which it would not get if the battery was this weak.

No, the inverter needed 10.5 according to the OP.

Nothing, because as long as you keep this up, you are being a loser.

You worked in a muffler shop on a bending machine.

And you can back this up how???? Oh, that's right, you can't. All of the evidence, including the very load test you insisted on say you are wrong. Get over it.

LOL, Mopars were also known for making some of the best automatics on the planet and yet you are whining about one of their screwups there as well.

LOL, what in the hell are you talking about.

Yea, but it is due to a fault in the wiring, not the battery.

And wrong.

Which has nothing to do with anything at all.

And sometimes those without the education think they know everything and then demonstrate the opposite.

Oh yea, Chrysler never screwed up. I guess that's why they never got above #3.

I recall it as being 11.5v. Given what its used to feed, I'm surprised its not higher.

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.

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

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.

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.

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

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')