Thanks for the reply, Willbe. I'll try to go thru those calcs to see if I can make heads or tails of them.
In the meantime, here is what I have done so far:
When I Googled "what is the resistance of 26 awg stranded copper wire?", I ran across this result:
https://www.powerstream.com/Wire_Size.htm
They have a tool to calculate the:
Voltage drop
Voltage at load end of circuit
Per Cent voltage drop
Wire cross section in circular mils
You just need to enter or select:
Copper or Aluminum
American Wire Gauge (AWG) Size
Voltage
1-way circuit length in feet
Load in amps
I have no idea what to enter for the load, but when I entered 3 Amps, it told me the voltage at the end of the circuit would be only 5.7 volts with a voltage drop of about 52%. That didn't sound right, but I don't really know. I just didn't think I could fire the ignitor with the voltage being that low.
Since I am reading an old textbook on "Basic Electronics by Grob," I thought I'd try out some of what I have learned so far.
The wire I am using is listed as 426-DTV, which is described online as 4 stand, 26 AWG stranded, tinned copper wire.
I measured the following resistances:
ignitor = .4 ohms
23 ft of 26 AWG = 0.7 ohms
46 ft of 26 AWG = 1.1 ohms
Using 40.81 ohms per 1,000 ft of 26 AWG wire, I calculated the following:
23 ft of 26 AWG = 0.94 ohms
46 ft of 26 AWG = 1.87 ohms
So I guess the total resistance from the firing box to the ignitor and back, would be between 1.5 and 2.27 ohms, and when all connected it measures 2.1 ohms.
The battery voltage measured 12.73 volts at the terminals. When I connected the 23 ft cable and measured the voltage at the unconnected end leads, it still measured 12.73 volts.
Using the measured resistance of 2.1 ohms, I = V/R = 12.27 / 2.1 = 6.06 amps. However, the 30 AWG steel wire fuses at 3.15 amps.
This is where I gave up due to spinning head.
============================
This reminds me of my first job in a computer shop. We were running old DG Nova 1220's that had a lot of 74xxx components. The first time one of these failed, the department head had me disassemble it and I was so excited that I'd get to watch him hook up the scope and troubleshoot it. After I had it all taken apart, he came over and pointed to some 74xxx chip and told me to pull it out, install a socket and plug a new chip in. After I had done that, he looked at my work and told me to put it all back together. It still didn't work, so he told me to do it all over and he pointed to a different chip to replace. I went through this about 4 times, sometimes just adding a cap here or there, before I finally asked him why he wasn't going to hook up the scope and troubleshoot the circuits to find the problem, He told me that he might spend hours and hours chasing down the problem and not find it, but in the long run it was cheaper for him to have me to go in and replace quite a few chips or add a few caps, than it would be for him troubleshoot the problem. Frustrating at first, but over time I got to understand why he suspected certain chips or signals and could do a pretty good job what I called "seat-of-the-pants" repair on those boxes.
In this case, from those numbers above, it seems to me that I shouldn't be able to fire a 1 inch piece of 30 AWG steel wire at the end of a 23 ft 26 AWG lead with 12 volts without melting the lead wire. But, so far at least, it never fails to fire and I can see no evidence of any damage to the leads. I just hope that someday I can calculate all this stuff.
Thanks again for your post.
Ken