1998 Grand Caravan

Complete and utter nonsense. The starter is completely current dependent and if a small load of around 10A can pull it down so quickly, what do you think the starter, pulling around 20 times that load is going to do to the voltage of that battery. Yea, those injectors will fire real well at around

Again, wrong. All that needs to be done is to measure the voltage at the battery under load which is NOT the same thing as a load test.

While true, we already know that the battery is not the problem.

Measured battery voltage under the load of the device.

Nope, it is not. Not enough current for a valid LOAD test or even enough time according to you.

And it wasn't a load test either. Jeez, what an idiot!

You always load test the battery because you don't fully understand what you are doing and it seems to work. Pretty much like blood letting reduced fevers so they did it out of habit without actually understanding how or why.

Simple yes, necessary here, not at all!

Except that no matter how you spin it, at very least a cobbled load test was done. YOU know it, because you made reference to it.

simple enough a length of wire is cheap and easy to find.

Aside from that, it would

yes it does max. if you have a load connected so the circuit is complete then measuring from the outer shell to neg post should result in 0 volts (or nearly) same with positive post to center post (you'll have to back probe as a load needs to be connected to complete the circuit.) now using ohms law we know that if there is resistance in either of the pos or neg side of the circuit you will see a "voltage drop" meaning you will read a voltage acrost that resistance. if the voltage is 0 then that means the resistance is 0.

Right, so are you measuring the voltage or resistance in the wire and its cobbled connections or the actual circuit you intended to check?

Right, but you did not say that before, so now you are doing a load test, something that the "experts" say isn't necessary.

Thats great, but the "negative" side won't have any power. So testing for voltage will net you a zero. If you wish to test the ground side for continuity to ground, a resistance check is in order.

Fact is, there are about ten ways to check each side of the circuit. The best way with the opening facts given is to check your power source for voltage and voltage under load, and see how it compares to your power port. That will eliminate the battery or the wiring. IF its the wiring, as is widely suggested (what engineer would allow THAT??) then a continuity check on the ground side is in order. If that proves to be good, then rewiring the port is the next step.

so there is not power on the ground side of a connection? funny, ive found plenty of voltage drops in corroded "ground" wires and connections before with exactly this test. try it with a resistor and a lightbulb from radio shack someday.

On a power port with nothing plugged in, there is no power on the ground side.

If the ground wire is to connected load which is in turn connected to a power source, that is correct. In the case of a power port with nothing plugged in, you will get NO voltage reading on the ground side.

I would, but without a battery, I know what I would find. NO voltage. This circuit would only have voltage if you had a battery or Uncle Fester connected to it.

Isn't it amazing how the same "bad" battery is supposed to drop enough voltage to trip the 10A inverter off, fail a load test (by that, I mean the true load test that you're talking about - not the super-light load test of the 10A inverter), and yet start the engine with no hint of a problem? I'd like to see the math on that one.

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

Damn! Here is what you said: "However, it does not determine if the voltage drop under load is due to wiring or battery."

Then Christopher: "yes it does max. if you have a load connected so the circuit is complete then measuring from the outer shell to neg post should result in 0 volts (or nearly) same with positive post to center post (you'll have to back probe as a load needs to be connected to complete the circuit.) now using ohms law we know that if there is resistance in either of the pos or neg side of the circuit you will see a "voltage drop" meaning you will read a voltage acrost that resistance. if the voltage is 0 then that means the resistance is 0."

Once again, you are proving your lack of understanding of Ohm's law and your dishonesty in the discussion (or intelligence - again, sometimes it's difficult to separate the two).

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

Christopher, Max and Budd stop just short of saying that Ohm's law is false, and it is clear that they do not understand it's application. Plus the fact that Max, thru dishonesty or ignorance plays shell games with the facts

- for example - here you clearly say that to do the voltage drop thing, you apply a load, and in Max's post immediately following, he says that you won't get a voltage drop because you didn't apply a load.

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

yes i know thats why ive talked myself out of replying serveral times. ive followed the thread for the most part from the begining.

I already went through how it could happen, and often does. Take a look at back posts.

Once again, when someone changes their description of the test, it changes the method of testing. Chris changed from simply probing the port to back probing the plug of the device plugged into the port.

Two entirely different things get two entirely different results.

No lying, except on your part, where you fail to observe the difference in Chris' posts.

Yet another lie on your part. Go back and read the original description Chris gives of his test.

Ohm's law works just fine Chris, its your testing methods that seem to change once I point out the flaws. Or is it just the description because you didn't actually describe it correctly?

Whichever, I'll allow you two self proclaimed geniuses to your flawed logic and devices.

Here it is: With a 10A load, we assume a voltage drop of approx 1 Volts. Thus, we can approximate the internal resistance at .1 ohms. Now, when starting the car, we can assume a current of 250A, which would lead to a voltage drop of 25V. Since this is greater than the nominal 12V output of the battery, we actually have a battery voltage of -13V (12V - 25V). -13V is clearly sufficient to start the car.

Even though the voltage is negative, because the current passes through both stator and rotor windings, the starter always turns in the same direction.

That was easy!

Hey, think you made one tiny mistake here...

Every time I've done lt the starter motor has turned backwards, which causes the engine to run backwards.

Have to go downtown in one speed reverse quite slowly, trying to go minus 50 kmh. It's OK when I get home, though, cause backing up the driveway in Drive is really, really efficient and fast.

Oh, and one other thing to be careful of. Watch your temp guage, 'cause the fans turn backwards, too, and the durned rad freezes solid.

The good thing is the radio runs backwards too, so instead of pushing out sound it sucks it all in. Nice and quiet for a change.

Ken

Ha ha! I love it! Ingenious! :)

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

I hear ya! I swore to myself that I wasn't going to reply anymore. I'm so weak!!!!!! :)

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

i never changed my post or method. read again max