Grounding confusion

Ignore all the audio equipment for a moment.. reduce it to the basics. You have a V of 14.4vdc, a circuit resistance of 0.5 ohms, apply ohm's law and calculate current.

Voltage = amps * resistance

14.4 = ? * 0.5

14.4 / 0.5 = 28.8 amps

Looking at the simulations that stevdart linked; 500 feet of 10 g is approximately 0.5 ohms. Enter those numbers into one of the wire points and see what happens to all the numbers.

Until you consider that the 0.5 ohm reading is incorrect, Ohm's Law states that you can't push 100 amps with 14.4 volts through a 0.5 ohm resistance.

http://ourworld.compuserve.com/homepages/Bill_Bowden/ohmslaw.htm

Jim

I showed that it is illogical to think that a mere 1/2 ohm of circuit resistance is going to reduce a car's sound sytem, no matter how big it is, to less than 300 watts.  And I showed that it normally takes a series of Ohm's Law equations, not just one, to find an answer.

haemphyst explained it very well, I thought.  I just don't know the actual calculations he used to find the loss value.

If you would, please explain when you use the simulator you provided

http://www.h-o-alternators.com/TechCalcs/voltdropthroughsystemhoalt.swf

and change the specs of the ground wire on the battery to 500 feet of 10 gauge wire (to simulate a 0.5 ohm resistance)

Move the volume to show an amplifier power without losses of 250

and read an amplifier power after losses of -27, and a total voltage loss of 15.275v, and a voltage at the amplifier terminals of -1.475v. Moving the volume slider around, the best output I can get is 56 watts. That's with an amplifier power without losses of 100+.

Explain to me what I'm doing wrong. I used your simulation, I added 0.5 ohms of resistance on the ground path, and the system doesn't work.

Jim 

I don't know how Haemphyst calculated it either.. I'd be interested in see the calculations. If I was doing calculations for sizing that resistor wattage I'd do something like;

watts = volts * amps

A 0.5 ohm resistor with 100 amps pushing through it is going to be dropping 50 volts.

watts = 50 volts * 100 amps

watts = 5000

Jim

I can't answer that because you are going on the presumption that the Ohm's Law formula you used is correct for the situation, as you are equating 1/2 ohm circuit resistance to 28 volts (the power/loss demo equates this long wire to a loss of 28 volts).  Actually, you were referring to 28 amps previously...

That's the number (28) you were looking for when you arrived at 500 feet of 10 gauge wire in that application.  I'm not studied in the electrical field, but I'm a logical person.  I have measured loss in wiring and circuits...and I use my logic to conclude that 1/2 ohm resistance through the chassis doesn't equate to a whopping 500 feet of 10 gauge wire.

This is just a question I've never seen pop up, maybe because the premise of it is so far-fetched.  The original author of this thread saw for himself that there was little, if any, discernible difference after he upgraded the ground.  I have a lot of studying to do to find the actual reason that you and he believe that 1/2 ohm circuit resistance makes such a major impact.  But I can see that you are still using this resistance factor alone in the equation.  You have to consider this:  you said to forget the sound system in this.  But this is the point:  without the amplifier(s) there isn't a load and consequential demand for amperage, and therefore no circuit.  Without a circuit there is no voltage drop or loss due to circuit resistance.

I think there has to be a series of equations done, not just one.  I just don't know what they are yet.  And the only thing I can pinpoint that is wrong with your use of the Ohm's Law formula is that you are leaving out the load on the circuit altogether.  Even if you add this 0.5 ohms as a resistance factor to the load, the answer doesn't seem to be right.  For example, a 2 ohm load with 1/2 ohm added to it makes a much bigger difference than a 4 ohm load with 1/2 ohm added to it.  One formula, W = I^2 * R (power loss in watts increased as the square of the current) may have something to do with solving this but I'm not sure how to use it. 

I can assure you that I'm not equating 1/2 ohm circuit resistance to 28 volts.. the power/loss demo is flawed. If you take the demo up to about 230 amp without loss you'll see the correct numbers; the voltage available to the amp is around 0.2v and the rest of the voltage is dropped across the 1/2 ohm resistance. It appears that the demo doesn't take into account increased circuit resistance and modify the current accordingly. You can see evidence of this by changing the cable length of the ground cable from 5 feet to 5000 feet and not seeing a change in voltage drop on the other cables, this indicates that circuit current isn't changing.

http://www.powerstream.com/Wire_Size.htm

I linked the chart that gives ohms per 1k foot numbers.. 500 feet of 10 gauge is 1/2 ohm. So is 100 feet of 17 gauge,  and 50 feet of 20 gauge. And if you enter each of combinations into the demo, the results are pretty much the same. I'm taking the chart as fact. If you can find something different, let's see it

I imagine that the reason the OP didn't see much difference was at least partially because he wasn't using a meter accurate enough.. as I mentioned earlier in the thread, I have a number of high dollar meters, and I don't feel they are accurate enough for this particular measurement.. that's why I always suggest voltage drop measurements.

When I said to forget the sound system in this I meant to ignore the audio side of it.. don't worry about trying to calculate power usage by measuring output etc, etc. For an example, let's say the amp draws 100 amps WFO. We can calculate apparent resistance by dividing 14v by 100 amps. To the battery, the amp looks like a 0.14 ohm resistance. If you add a 0.5 ohm resistance to the circuit, then total circuit resistance becomes 0.64 ohms. Maximum circuit current then becomes 14 volts divided by 0.64 ohms, or 21.87 amps. You'll never see more than 21.87 amps in that circuit.

http://www.angelfire.com/pa/baconbacon/page2.html

For watts, use Watts = volts * amps.. it's much easier.  WRT your 2 ohm 4 ohm examples, are you talking about audio or power? A 1/2 ohm increase to 2 ohms is a 25% increase, with 4 ohms it's only a 12.5% increase.

Ohm's Law is easy.. you're way over thinking this. The best way I could demonstrate this here would be to recommend to you to get an old style headlight.. one of the large retangular ones would work good. Ohm it out; it'll probably be close to 1/2 ohm or less. Wire the headlight into the power circuit of one of your amps.. you could probably open the circuit at the fuse and install it there. You've just introduced a 1/2 ohm resistance into your amp circuit.. see what happens.

Jim  

Correction.. a standard 60 watt headlight is going to be around 2 ohms.. so parallel a couple to get 1 ohm if you'd like.

Jim

Personally, I think the reality is that the measurement isn't accurate.. but if you were so inclined you could do the 'hook a headlight into the amp power circuit' trick I mentioned earlier in the thread. That'll put a 2 ohm resistance into the power circuit; I promise you'll see the difference.

I  have an electronics degree,  and 20+ years experience, more than half of which is automotive electrical/electronics; ASE certs and all that. Currently I work for a state agency doing electronics, and moonlight for a local university doing, among other things, systems automation.

I'll be the first to admit I don't know it all.. the more I learn, the more I realize how little I know.

But I do know a little.

Jim

Well.. if you have any questions, just ask. Working purely in theory makes it more difficult, which is why I suggested the headlight mod; hands-on beats theory every time from an understanding perspective.

Jim