1998 Grand Caravan

Given that removal of the battery is not necessary,

No, it is not. For as educated in this as you all claim to be, you don't
seem to realize that a load tester is easily found and under $50. Thus many
places have them, and loan them out. All you need to do is clamp them on and
push a button, no need to move the battery anywhere.

An example seen here:

http://www.toolsource.com/ost/produ...6095&mscssid=X0JEU6MD7P9K9KLGDBJM2VJPUSAM2889

1) See my reply to Bill regarding the very real possibility that the starter
works fine while the power port may not.

2) The only way to determine if its the wiring or the battery is to load
test the battery.

3) testing the ground could easily be more difficult than testing the
battery.

Alright, if you plug in the inverter, and you place a load on the battery
via the port, what have you done? Yup, thats right, LOAD TESTED the battery.
And on top of it all, you haven't removed the battery. Jeez, that was tough.

Yup, thats why I prefer to load test the battery. Then I can see how the
battery works under load without all the rest of that stuff as variables.
Testing the battery determines a baseline, and allows for proper testing of
the rest of the varibles.

Terrific, see above regarding the ease of each test and why a battery load
test is so simple.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

Even easier. with the side benefit of automagically identifying

This would be effective if we didn't already know there was a difference in
voltage via the inverter function.

You'll find nothing here, since both are grounded. Assuming the socket is
losing its ground, you'll still find nothing. If you measure the resistance,
THEN you'd find a bad ground.

The downside to these tests is the fact that most multimeters do not have
leads long enough to complete the desired circuit. Aside from that, it would
be a very effective method of checking the circuits. However, it does not
determine if the voltage drop under load is due to wiring or battery.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

--
If at first you don't succeed, you're not cut out for skydiving

And you will get NOTHING. Why, because both sides are connected to positive
and with no load, there can be no drop for the meter to measure.

Again, if it were the socket that were bad you would still get nothing
because there is no current flow there when the socket is not being used.
If you have a bad body ground, you would see a voltage but it would not tell
you where the problem is. In this case I would do as Max suggested and
measure for resistance.

Both are probably going to be zero because without a load, you don't get
voltage drops.

 

http://www.toolsource.com/ost/produ...6095&mscssid=X0JEU6MD7P9K9KLGDBJM2VJPUSAM2889

Damn Max, are you ever going to let this one die. There is no need for a
load test here, no matter how hard you try and spin it.

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
8 volts or less.

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.

tough.

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.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

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.

 

The downside to these tests is the fact that most multimeters do not have

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.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

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.

 

so there is not power on the ground side of a connection?

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.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

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.

 

Isn't it amazing how the same "bad" battery is supposed to drop enough

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

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

Once again, you are proving your lack of understanding of Ohm's law and

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.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

Max and Budd stop just short of saying that Ohm's law is false, and it is

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

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

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.

--
Max

"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author)

 

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')