posted 04-07-2006 07:31 PM ET (US)            

The good news is the problem is reproduceable.

I'd start by eyeballing all the connections in the console, and loosening
all the screw connections a half turn, and retightening, and
pulling all the slip on connections off and back on again.

Get a digital voltmeter.

Step 1.: Turn everything off. Measure the
voltage across the battery, and I mean across the battery --
touch the battery posts, not what's clamped on them. It
should measure a bit over 12V. If it doesn't you have bad
battery or charging system. Then turn everything (FF, GPS,
lights and wipers) on. It should read just a touch (.1V?)
lower. If it doesn't, you have a bad battery.

Step 2: Find where power comes into the console (or dashboard).
Measure power with everything off, and everything on.
Results should be as in Step 1. If they aren't you have a
bug in the wiring between here and the battery.
I'd bet even money or better you will have found the bug
by now.

Step 3: measure where the GPS and FF are wired in, everything
on and everything off. It should be as in Step 1, if it
isn't, the bug is between here and Step 2 (Aux switch would
be the first suspect). Odds are about 100% you've found
the bug.

And BTW, you probably don't have a short (a connection where
there shouldn't be one), but rather a place of high, but less
than infinite, resistance.

Also, pull the fuses and put them back. I suspect the lights
and electronics and wipers are on different fuses, but it's
still quick and easy and worth a shot.

posted 04-08-2006 09:57 AM ET (US)            

The voltage drop across a resistance is directly proportional to the current flow:

E = IR

where

E = electromotive force in volts
I = current in amperers
R = resistance in ohms

i.e., Ohm's law.

If a bad connection exists in your 12-volt distribution wiring, it adds resistance. Assume it adds 5-ohms of resistance.

The current drawn by the electronic devices is small. Let us say 0.2-ampere. The voltage drop across the bad connection will be

E = IR
E = 5 x 0.2
E = 1.0 volt

Thus the 12-volt supply voltage to the electronic devices will drop to 11-volts. This is still enough for them to operate.

When the navigation lighting circuit is energized, the lamps draw about 2-amperes of current. The voltage drop across the high resistance will now be:

E = IR
E = 5 x 2.2
E = 11 volts

Now the supply voltage drops to 1-volt. The electronics shut off.

This example is a bit simplistic, but it demonstrates the problem.

Also, if the above situation were to occur, consider the power being dissipated across the bad connection.

Power is proportional to current and voltage:

P = IE

By substitution and algebra, we can combine the two formulas to solve for power in terms of voltage and resistance:

P = IE and E=IR, or I=E/R
thus
P = E(E/R)
P = E^2/R

If 11-volts drop across a 5-ohm load, the power dissipated will be

P = 11^2/5
P = 96.8 watts

This is a great deal of heat and will soon make itself known.

In actual practice this won't be quite as described above. The current drawn by the lamps will be reduce by the drop in voltage supplied to them. For example, if each lamps is rated as a 10-watt lamp at 12-volts, this implies that three of them will have a resistance of

R = E^2/P
R = 12^2/(10 X 3)
R = 144/30
R = 4.8 ohms

If there is a 5-ohm resistance in the distribution, the 12-volt supply will divide across them in proportion to their resistances, thus across the lamps will be

E = 12 x [ 4.8/(4.8 + 5) ]
E = 5.87 volts

and the voltage across the 5-ohm resistance will be

E = 12 x [ 5/(4.8+5)]
E = 6.12 volts