Date:         Tue, 22 Jan 2008 18:41:47 -0800
Reply-To:     Pensioner <al_knoll@PACBELL.NET>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Pensioner <al_knoll@PACBELL.NET>
Subject:      Electrical stuff
In-Reply-To:  <200801230155.m0N1tTut009438@nlpi004.prodigy.net>
Content-Type: text/plain; charset="iso-8859-1"

Mike mentioned:

>>When you have poor, corroded wiring/ connections or bad grounding, voltage
drops.  When voltage drops, current rises automatically in proportion.  This
is due to the current-consuming devices demand on the circuit.

Not really.  The battery/alternator voltage remains the same.  The voltage
drop across the corroded wiring/connector rises leaving less voltage to be
applied to the load devices in the series circuit.  The current actually
decreases depending on the characteristics of of the load devices (linear vs
non linear)

>>An electric motor normally powered by 12 volts and normally drawing 5 amps
of current flow when running at full speed, will now draw 10 amps of
current, if the input voltage get reduced to 11.0 volts, in an attempt to
continue running at normal speed.  The wires get warmer as current flow
exceeds the designed capability of the conductor's diameter.

Never experienced that self controlling phenomenon in DC motors.  A/C motors
(synchronous) draw more current as the power factor changes from their
design specification, but DC motors just run slower.

>> The factory has designed the wire guage size to allow a small buffer to
allow the fuse to blow due to sudden, severe overload, before the wire would
melt and cause a fire, but not enough to allow for such a constant smaller
overload.

No detailed knowledge about this but it is merely your conjecture.  Do an
analysis of voltage drop over distance for various AWG conductors. (Google
is your friend here) and it may give you new insights into the 'wire heating
phenomenon' and what you might expect to measure.

>>Continuously running the circuit at this higher current flow will heat the
wire warmer than normal, and the wire insulation, plastic plug connectors
and switch bodys will eventually soften and melt from the excessive heat.

The cause is the increased resistance at the connector/switch contact not
higher current.

>>  For example, the Vanagon headlight switch routes all of the lamp's power
through it (instead of through relays), so as the vehicle ages and the
grounds, etc, become bad, the metal high and low beam contacts inside the
plastic switch housing get hot enough to melt the plastic rocker inside of
the switch, burning it up.

Exactly, the degraded switch contacts heat up because of the resistance of
the contacts.


>> When you replace the blown fuse, bad switch, or
melted connector, etc., what have you done to diagnose and fix the actual
problem?  Things melt for a reason, not just because they're old and worn
out!  The headlight switch doesn't melt because it's old, but because the
connectors and grounds are drawing excessive current.

Sorry, simply not so.  The current doesn't rise, the voltage drop across the
contacts increases and the resistance dissipates more heat with the added
resistance. Use a good ammeter that will register 10A and try it out.


>>  The Vanagon radiator fan motor, fan temp switch and their plug
connectors
sit outside in a sometimes wet, salty, corrosive environment for 15 to 25
years.  Even if they were weather-tight when it rolled off the showroom
floor, I bet that's no longer the case.  You could take voltage and amperage
measurements at the motor connection (while it's running) to determine
what's really going on there.  Measuring closer to the motor is better,
because the voltage will drop a little bit at each connection, and you need
to know what the final figure is getting down to at it's lowest point
farthest along the way, but still before the motor.

You're getting closer.  Measurement is the key.  Measure the current in the
loop, it will be a constant value no matter where in the loop you do the
measurement.  Now you know the current.  Now measure the voltage across the
switch.  The power dissipated in watts is the product of the Current and the
Measured Voltage across the switch or connection.

Substitute a known good switch and re measure both the current and the
voltage drop.

'I dunno, Leroy, ain't nobody I know ever seen an electron anyhow'

Maxwell said it all...