QUOTE(r_towle @ Apr 29 2023, 02:28 PM)
QUOTE(Superhawk996 @ Apr 29 2023, 03:04 PM)
QUOTE(r_towle @ Apr 29 2023, 01:35 PM)
the old wiring we have gets corroded inside, thus can no longer move 20 amps...and the starter may pull 30+ amps...which can Melt the wire somewhere you cannot see, or melt the ignition switch.
Rich
I’m sorry to have to point this out, but a circuit that was previously designed to draw 20A (as in example) will not suddenly draw 30A due to an increase in wiring resistance increasing from corrosion. This is physically impossible as dictated by Ohms Law.
However, I’m in rabid agreement with Rich’s suggestion to 1/2 split the system by basically hot wiring the starter B+ to the yellow solenoid control. Whether you do that with a screwdriver or a remote starter switch doesn’t make a bit of difference in the end. You then know if the problem is in the starter solenoid assembly or in the wiring and can focus on the appropriate solution.
This is why contribute our suggestions in the forum to help others correct us.
Its also why (DM me or give me a call) is not the way to help others.
a 20 amp circuit design will indeed melt wires that are designed to support 20 amps.
I do understand ohms law...but I was trying to explain things in a public forum , of non electrical engineers, to help.
Happy to hear criticism, but I am also quite comfortable in my own skin to help others in a public forum, so others can refine the answer and create a positive outcome.
Rich
As am I. So let’s help others by understanding what is going on.
So let’s be clear that a circuit designed to pull 20A will not melt when pulling 20A through good wire. It can actually melt when pulling LESS than 20A through degraded wiring.
How?
Let’s start with ohms law and a 12v system
To draw 20A, how much resistance was in the circuit as designed?
12v / 20A = 0.6 ohms of resistance
This 0.6 ohms of resistance would represent normal solenoid resistance + the wiring resistance which should be almost, but not quite equal to 0.
So now let’s say the wiring has corroded and increased to 0.5 ohms of resistance by itself. We now have a circuit with 0.5 ohms of wiring resistance + 0.6 ohms of solenoid resistance. This is 1.1 ohms of total resistance.
So how much current will we pull?
12v / 1.1 ohms = 10.9A
Circuit is now unable to pull full current. In fact, it is only pulling slightly more than 1/2 of the design current. So if in this example, the solenoid needs 20A of current to operate the solenoid, it can’t pull that current and the starter solenoid won’t energise properly. This will be true regardless of how many grounds are attached to the solenoid.
So why does the wiring melt when pulling less than 20A it was designed to pull?
The power (I.e heat) the circuit has to to dissipate is Power = current ^2 (squared) * Resistance
When wiring is near 0 ohms, there is no heat created . This is the design condition of healthy wire.
But we have 0.5 ohms of resistance, internal to the wire, created by corrosion.
So how much power (heat) will be created by pullling that 10.9A though the corroded wire?
Power =10.9A * 10.9A * 0.5 ohms
Power = 59 watts
So now we are asking a wire to dissipate 59 watts of heat. Remembering that it was not designed to do this. 59 watts may not sound like much but it is quite a bit. Trying to pick up a 10 watt ceramic resistor that is operating normally will be quite hot to the touch and uncomfortable to hold. We are talking about roughly 6 times hotter.
So this is how wiring melts from corrosion while actually pulling LESS current than what the circuit was originally designed for.