stevdart wrote:
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... |
|
|
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.
stevdart wrote:
Mad Scientists wrote:
you are correct; with 14v and a resistance of 0.5 ohms, the circuit current would be 28 amps. |
|
|
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. |
|
|
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
stevdart wrote:
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 not voltage drop or resistance loss. |
|
|
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.
stevdart wrote:
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 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 increasing as the square of the current) may have something to do with solving this but I'm not sure how to use it. |
|
|
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