Re: ignition and calculations

[cookiedan@domain.elided (Daniel Nees)]

John,
   I screwed up and didn't explain myself properly. The calc that I was
getting at was the amount of power the resister uses. Ex. 1 12V*.43
ohms=5.6V Now in your example of the 12V*2ohms=24V is corect as far as
the math goes, but; wouldn't work. Ex. 2 take an 120V light bulb and put
it in a lamp. You get a nice brite light from this. Now take that bulb
and apply 100V through it, you get light but it is dimmer. Now if you
took this same bulb and applied 50V to it it would barely light, and
again apply 12V to it and you wouldn't get anything through it. Why?
Because the lamp or resistor is made to produce light/heat with 120V. It
will still make light at 100V and maybe even 50V but as you get
progressively lower there is insufficient voltage to overcome the
resistance of the filiment in the bulb. So, 12V*2ohms=24V really is a
statement of what it would take to get through the 2ohm resistor. The
formula is for Voltage drop. E*R=Ed. This is how you would figure out
what size wire to use from the battery to your trailor. Take the
resistance of a size of wire (this is found in the back of the national
electrical code book per foot) and multiply that resistance by your
voltage. When you have a wire size that has the most acceptable lose for
the length of wire you are using then you use that size wire.
   I know there is more to be said on that thought, but; I'm drawling a
blank all of a sudden. But, I'm sure when you reply to this post I'll
relize what I forgot and say it.

Dan Nees
cookiedan@domain.elided

   1979 Scout II 345, Auto, 3.07's, Trak-Lok. One Ugly and Trusty Truck
covered in mud! Thinking about throwing a couple of big buckets of mud
on it just before I get to the fairgrounds!

1971 Scout II 304, D30/44, D20, power steering.

http://members.tripod.com/~IHCaholic/scoutindex.html