Date: Sat, 2 Mar 2002 10:07:13 -0500
Reply-To: David Beierl <dbeierl@ATTGLOBAL.NET>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: David Beierl <dbeierl@ATTGLOBAL.NET>
Subject: Re: Relaying Headlights
In-Reply-To: <016801c1c1c4$47b91ff0$0100a8c0@MEDION1800>
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At 03:28 AM 3/2/2002, Robert Steven Fish wrote:
>I also do NOT WANT to run all that wire... but I will do it if someone can
>tell me (with some good logic behing their arguement) that there is indeed
>an advantage to powering right off the alternator (sounds a bit like a
>THREAT, doesn"t it?)
Dinna bother, there's not any -- or not much. If there is insufficient
voltage at the battery, it's an absolute indicator that you need thicker
wires or better connections from alternator to battery -- or that you are
simply overwhelming the alternator by using more energy than it can supply
-- a common problem with ambulances and police cars.
Actually the most cost-effective thing you can do is modify the regulator
to use an external sensing lead, and run a 16-ga or so wire from there to
the battery. This will cause the alternator to develop whatever voltage
will result in the correct voltage *at the battery* even if it loses two
volts along the way (which is grossly excessive and would waste 12% of the
alternator output, but I'm making a point).
>But it is true... the alternator feeds the starter, and then heads right on
>up to the battery positive post. So maybe the battery post is, at that
*While charging -- the usual state when the engine is running* the
alternator is indeed the "highest point" in the system and everything runs
downhill from there. The rule, Ohm's law, governing voltage and current
relationships in a DC system is E (electromotive force, or voltage) = I
(current) * R (resistance). So if there is a resistance of 0.005 ohms
between the alternator and the battery, and supposing the alternator were
supplying 100 amps, the voltage drop along the wire would be 100 x 0.005,
or 0.5 volt. If the alternator were to supply only 50 amps, the voltage
drop would be 0.25 volts.
Also by Ohm's law (the Secret Policeman's Other Law), P (power) = E * I =
E^2 / R = I^2 * R. If you are dropping half a volt in a circuit passing
100 amps, that means that you are spending 50 watts or about one-fifteenth
horsepower to keep the bottom of the van warm. If you drop half a volt
across a connection, at 100 amps you have a handy little 50-watt soldering
iron, which is why the insulation melts when your crimp connections go bad.
>point distributing all the charge to the rest of the car, while also sucking
>juice to recharge, and cannot provide the desired voltage?.... I have no
>idea.. but would love to find out what the real deal is.
The battery (to the rest of the system) looks roughly like a gigantic
capacitor, smoothing out the voltage pulses from the alternator so that the
system sees a fairly steady voltage instead of a rapidly pulsating one.
Run your lights from the battery. If the voltage there isn't enough,
improve your wiring between alternator and battery, use external sensing on
the regulator, or both.
david
--
David Beierl - Providence, RI
http://pws.prserv.net/synergy/Vanagon/
'84 Westy "Dutiful Passage"
'85 GL "Poor Relation"
