So what you're saying is no need to turn in those old batteries, just hook
up a new one in parallel ... and oh yeah, no problem running a starting
battery (my understanding that's what GC has) for accessories until it's
dead ... they can be recharged over and over and over? Geez ... even a deep
cycle's life is shortened by doing that. I do agree that a healthy lead acid
fully discharged has a very high internal resistance ... as a matter of
fact, so high that it cannot be charged using the alternator ... my jeeping
buddy found that out. But about that old starting battery (that GC has)
that's maybe dry to the plates or maybe bridged or ... my quick and dirty
risk assessment ... adios old batteries ... I for one would do not connect
an old battery in parallel with my only way back to shore. So post us a
picture of your batteries configuration ... I'm always up for a good laugh
.... adios ... "killer" :-)
"Folklore killer" wrote in message
...
"bowgus" blathered....
in Message-id: ...
"GC" wrote in message
...
I'm about to wire up a battery switch so that I can switch between two
batteries. (one or the other or both)
Don't use the "both" setting ... read on
Just wondering if I should put the switch on the positive or negative??
Or
doesn't it matter?
Positive is the norm around here (geez ... hurts the head to even think
about switching the negative since that's normally the electrical
reference)
Also, do both the batteries have to be the same age/rating??? I was
hoping
to use my old battery for powering accessories when I'm stopped, but
charge
it when the engine is running with the other battery.
In your situation, I would ... (get myself a deep cycle for the
accessories
... e.g. my accesory is my electric trolling motor)
a) make sure the switch is a make before break type which they normally
are
so that the alternator is never without a load if switched while running;
b) when stopped select the old battery (if it's a regular starting type
battery it won't tolerate many cycles of being discharged), before
starting,
switch to the new battery;
c) while running charge batteries by switching between the two (hmmm ...
when are they charged ... well, maybe your batteries have indicators of
the
green it's charged variety?);
d) never connect the batteries ... it's my understanding that the charged
one could pump a lot of current into the discharged (my theory ... a
battery (let's just say for the sake of argument rated 800 cold cranking
amps) has no current limiting feature like a battery charger does) which
could result in 800 amps going through wiring/switch resulting in fire,
exploding battery, ...
Thanks in advance! 
My 2 cents worth ...
And worth slightly less than 2 cents.
Where do you people get your information? Disneyworld?
A typical engine start battery would be 600 to 650 CCA
As you have quoted 800 CCA we shall use that as the example.
A brand new 800 CCA engine start battery will have an internal resistance
of
around 0.004 ohms and an _open circuit_ voltage of around 12.7 volts (you
can
calculate the fully charged internal resistance of an engine start battery
using the given CCA and the _definition_ of what CCA means - but of course
you
didn't know that).
A flat deep cycle battery will have an _open circuit_ voltage of around
11.9
volts and an internal resistance of around 0.06 ohms.
The voltage _difference_ between the 2 batteries is therefore 0.8 volts
dropped
across the two internal resistances of 0.004 + 0.06 = 0.064 ohms = 12.5
amps.
And that is making the simple assumption that the voltage of the fully
charged
engine battery does not fall under load (in addition to the voltage drop
caused
by the internal resistance) which in practice it will, by about 0.2 volts
at
this sort of loading.
So the _real_ current draw will be approximately 9.4 amps.
Further, there is a dead band between the open circuit voltage and the
voltage
at which charging will commence. This dead band varies quite dramatically
depending upon the state of charge of the battery in question. A typical
figure
for a flat deep cycle battery would be about 0.2 volts.
So an even closer figure for the current draw (that you assert is 800
amps)
would be (12.5V - 12.1V) 0.4V/(total internal resistance) 0.064R = 6.25
amps.
And of course you won't believe me. So instead of using half a theory that
you
don't understand anyway, either go learn the rest of the theory or try it
and
take some measurements.
Connectng the batteries in parallel is perfectly safe, quite normal, will
achieve what is desired and is used in millions of installations the world
over.
Idiot
Folklore Killer.