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30 May 2016 03:19:44 GMT in alt.home.repair, wrote:

http://www.uglys.net/

You have voltage drop due to the length and size of the wire. Because the wire isn't a super conductor.

At the end of the day, you might have put 120 volts on the line, but, 100ft down that line, you aren't getting 120 volts. Some has been lost on the wire, due to the wires own resistance.

Switch from AC to DC with no other changes, and the voltage drop is more pronounced.

On Monday, May 30, 2016 at 1:57:48 AM UTC-4, Diesel wrote:

Heh, smart guy, give us your formula for that voltage drop in the wire. The rest of us here, have told you that from a very simple application of Ohm's Law, it's V = I*RW where I is the current flowing and RW is the resistance of the wire. Set I to zero, what's the voltage drop? ZERO. So, we can calculate it. Show us YOUR calculation.

More FUD and diversion. The differences between switching from 240V AC to DC in this discussion is totally negligible. Another example of an amateur taking something they heard or saw and misapplying it.

PS: With DC, just like with AC, if the current is zero, the voltage drop on the wires is zero.

Why? Why not just do the sums and work out what's actually needed instead of believing some paper pushing fuckwit?

#### Site Timeline

- posted on May 30, 2016, 5:57 am

http://www.uglys.net/

You have voltage drop due to the length and size of the wire. Because the wire isn't a super conductor.

At the end of the day, you might have put 120 volts on the line, but, 100ft down that line, you aren't getting 120 volts. Some has been lost on the wire, due to the wires own resistance.

Switch from AC to DC with no other changes, and the voltage drop is more pronounced.

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- posted on May 30, 2016, 3:38 pm

Heh, smart guy, give us your formula for that voltage drop in the wire. The rest of us here, have told you that from a very simple application of Ohm's Law, it's V = I*RW where I is the current flowing and RW is the resistance of the wire. Set I to zero, what's the voltage drop? ZERO. So, we can calculate it. Show us YOUR calculation.

More FUD and diversion. The differences between switching from 240V AC to DC in this discussion is totally negligible. Another example of an amateur taking something they heard or saw and misapplying it.

PS: With DC, just like with AC, if the current is zero, the voltage drop on the wires is zero.

- posted on May 30, 2016, 10:11 pm

wrote:

Absolutely totally 110% wrong.. There is NO voltage drop withouit current flow.

You are WAY in over your head.

Absolutely totally 110% wrong.. There is NO voltage drop withouit current flow.

You are WAY in over your head.

- posted on May 30, 2016, 11:16 pm

On Monday, May 30, 2016 at 6:11:29 PM UTC-4, snipped-for-privacy@snyder.on.ca wrote:

We should also point out that it's more BS that the voltage drop would be more pronounced if we switched from AC to DC. It would be exactly the same, excluding negligible second order effects like skin effect, inductance, etc. And those produce the reverse effect, ie more resistance to AC current than DC.

We should also point out that it's more BS that the voltage drop would be more pronounced if we switched from AC to DC. It would be exactly the same, excluding negligible second order effects like skin effect, inductance, etc. And those produce the reverse effect, ie more resistance to AC current than DC.

- posted on May 31, 2016, 1:28 am

snipped-for-privacy@snyder.on.ca pretended :

That's what***I*** said, and you said I was wrong. Sheeesh, make up your
mind.

That's what

- posted on May 31, 2016, 4:03 am

On Mon, 30 May 2016 21:28:19 -0400, FromTheRafters

You appear to have said, I believe "You have voltage drop due to the length and size of the wire. Because the wire isn't a super conductor. "

You also appear to have said "At the end of the day, you might have put 120 volts on the line, but, 100ft down that line, you aren't getting 120 volts. Some has been lost on the wire, due to the wires own resistance. "

And the context was an "open circuit."

Under a load, you are correct. On an open circuit, you are just plain wrong.

If you did not say what you appear to have said, accept my apologies.

You appear to have said, I believe "You have voltage drop due to the length and size of the wire. Because the wire isn't a super conductor. "

You also appear to have said "At the end of the day, you might have put 120 volts on the line, but, 100ft down that line, you aren't getting 120 volts. Some has been lost on the wire, due to the wires own resistance. "

And the context was an "open circuit."

Under a load, you are correct. On an open circuit, you are just plain wrong.

If you did not say what you appear to have said, accept my apologies.

- posted on May 31, 2016, 11:26 am

snipped-for-privacy@snyder.on.ca submitted this idea :

No, that most assuredly wasn't me.

Nope, again not me.

Nope, not me. I responded to Al Gebra's:

"Voltage drop is represented by the formula E=I*R Seems to me if the current flow is zero, then the voltage would be zero as well."

With:

"That's a good theory, but IMO it is wrong."

I say it is wrong because "voltage drop" depends on the dissipation of energy by a device with current flowing through it. No current, then you are not even talking about "voltage drop".

If Al Gebra had said "voltage" instead of "voltage drop", or if nobody had argued about my opinion (stated clearly as my opinion) then much of this conversation would not have taken place.

Accepted.

Aside, do you think that Ohm's Law works for zero current or zero resistance values? I fully accept that it works for the case of zero voltage, but not for current or resistance because division by zero is undefined.

Thanks in advance.

No, that most assuredly wasn't me.

Nope, again not me.

Nope, not me. I responded to Al Gebra's:

"Voltage drop is represented by the formula E=I*R Seems to me if the current flow is zero, then the voltage would be zero as well."

With:

"That's a good theory, but IMO it is wrong."

I say it is wrong because "voltage drop" depends on the dissipation of energy by a device with current flowing through it. No current, then you are not even talking about "voltage drop".

If Al Gebra had said "voltage" instead of "voltage drop", or if nobody had argued about my opinion (stated clearly as my opinion) then much of this conversation would not have taken place.

Accepted.

Aside, do you think that Ohm's Law works for zero current or zero resistance values? I fully accept that it works for the case of zero voltage, but not for current or resistance because division by zero is undefined.

Thanks in advance.

- posted on May 31, 2016, 5:51 pm

On Tuesday, May 31, 2016 at 7:26:10 AM UTC-4, FromTheRafters wrote:

We can use Newton's Law to calculate the velocity of a ball thrown straight up in the air at 20M/sec along it's entire trajectory. At the peak, the velocity is zero. According to you, that would be wrong too, because without movement, there is no velocity. In science and engineering we solve these equations all the time, and zero values have meaning. Often the zero values are the most important ones.

Did you do that graph yet? Plot Ohm's Law. It's a straight line through the origin. The rest of us all see a solid line. You for some bizarre reason insist that there is some unique singularity at the origin, when there isn't. You insist that graph line isn't valid because there is some division by zero going on, when there isn't.

Some people's "opinion" is that the earth is flat too. That doesn't make it right and there is no issue of opinion here with Ohm's Law. Everyone is on the same page, except you.

Again with the division by zero. No division is involved with:

V = IR. I = 0 or R = 0, then V =0

No division by zero, idiot.

We can use Newton's Law to calculate the velocity of a ball thrown straight up in the air at 20M/sec along it's entire trajectory. At the peak, the velocity is zero. According to you, that would be wrong too, because without movement, there is no velocity. In science and engineering we solve these equations all the time, and zero values have meaning. Often the zero values are the most important ones.

Did you do that graph yet? Plot Ohm's Law. It's a straight line through the origin. The rest of us all see a solid line. You for some bizarre reason insist that there is some unique singularity at the origin, when there isn't. You insist that graph line isn't valid because there is some division by zero going on, when there isn't.

Some people's "opinion" is that the earth is flat too. That doesn't make it right and there is no issue of opinion here with Ohm's Law. Everyone is on the same page, except you.

Again with the division by zero. No division is involved with:

V = IR. I = 0 or R = 0, then V =0

No division by zero, idiot.

- posted on May 31, 2016, 6:17 pm

On Tuesday, May 31, 2016 at 7:26:10 AM UTC-4, FromTheRafters wrote:

That's like saying that when we have a velocity of zero in physics, we can't say that the velocity is zero, because we're "not even talking about velocity". Or when an object is at rest, it's momentum is zero, because you are not even talking about momentum. We can use Newton's Law to calculate the velocity of ball that is shot upward. We can calculate the velocity at every point. At it's peak, the equation gives a velocity of zero. According to you though, that would be wrong, because "we're not even talking about velocity". You didn't take a class in physics, did you? We worked out these problems all the time. And many times the value of zero not only has meaning, it's one of the most interesting.

That's like saying that when we have a velocity of zero in physics, we can't say that the velocity is zero, because we're "not even talking about velocity". Or when an object is at rest, it's momentum is zero, because you are not even talking about momentum. We can use Newton's Law to calculate the velocity of ball that is shot upward. We can calculate the velocity at every point. At it's peak, the equation gives a velocity of zero. According to you though, that would be wrong, because "we're not even talking about velocity". You didn't take a class in physics, did you? We worked out these problems all the time. And many times the value of zero not only has meaning, it's one of the most interesting.

- posted on May 31, 2016, 5:30 pm

snipped-for-privacy@snyder.on.ca

That was me.

I've already agreed that I was wrong.

That was me.

I've already agreed that I was wrong.

--

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

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- posted on May 30, 2016, 5:25 am

On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

[snip]

Where can I get one of these magic heaters, that uses 1500 watts when it isn't connected?

[snip]

[snip]

Where can I get one of these magic heaters, that uses 1500 watts when it isn't connected?

- posted on May 30, 2016, 6:14 am

Mon, 30 May 2016 05:25:23 GMT in
alt.home.repair, wrote:

Hmm? Oh.. ROFL, I see. That's what happens when you don't pay attention to context and choose to selectively quote.

Those are load calculations based on a 1500watt element running at two different voltages. It shows the amount of amps required to get 1500watts at those voltages.

The different voltages are directly related to length and diameter of wire and have nothing to do with the load from the heater at this point.

Thanks to the wire and the wire alone, some volts have already been lost (as in they'll never reach the heater) in transit, due again, to the wires own resistance and length. In this case, the length is the same; 100ft. But, the wire size or gauge is not. 12ga is on the top, 10ga is on the bottom; for comparison.

Might I suggest a simple english comprehension class or two? Muggles suffers from a similar problem. So, you're in very good company.

Hmm? Oh.. ROFL, I see. That's what happens when you don't pay attention to context and choose to selectively quote.

Those are load calculations based on a 1500watt element running at two different voltages. It shows the amount of amps required to get 1500watts at those voltages.

The different voltages are directly related to length and diameter of wire and have nothing to do with the load from the heater at this point.

Thanks to the wire and the wire alone, some volts have already been lost (as in they'll never reach the heater) in transit, due again, to the wires own resistance and length. In this case, the length is the same; 100ft. But, the wire size or gauge is not. 12ga is on the top, 10ga is on the bottom; for comparison.

Might I suggest a simple english comprehension class or two? Muggles suffers from a similar problem. So, you're in very good company.

--

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

Hmmm. I most certainly don't understand how I can access a copy of a

Click to see the full signature.

- posted on May 30, 2016, 3:48 pm

On Monday, May 30, 2016 at 2:14:05 AM UTC-4, Diesel wrote:

He's paying attention and his sarcasm is on point.

Except of course heating elements look like resistors and they don't magically readjust to pull more amps with a lower voltage to keep the watts constant. At 240V, a 3500W heating element pulls 14.6A. Following your logic, if we instead put 24V on it, it will pull 146 A and continue to put out 3500W. That makes sense to you?

Give us the formula for that voltage drop in the wire. The rest of us know it's V = I*RW, where I is the current, RW the resistance of the wire. Ergo, I = 0, V = 0. Capiche? That means that with no load, the full 240V is present at both ends of the long wire. Now put even a tiny load on and now you have voltage drop. The larger the load, the larger the voltage drop in the wires, because the current is higher.

Again obvious you don't understand the basics. Volts are not "lost" If they are and you're so smart, give us the formula for the "lost" volts with no current flowing.

Might I suggest a course in electricity 101? Understand how to apply] Ohm's Law to a simple circuit? Kirchoff's Laws?

He's paying attention and his sarcasm is on point.

Except of course heating elements look like resistors and they don't magically readjust to pull more amps with a lower voltage to keep the watts constant. At 240V, a 3500W heating element pulls 14.6A. Following your logic, if we instead put 24V on it, it will pull 146 A and continue to put out 3500W. That makes sense to you?

Give us the formula for that voltage drop in the wire. The rest of us know it's V = I*RW, where I is the current, RW the resistance of the wire. Ergo, I = 0, V = 0. Capiche? That means that with no load, the full 240V is present at both ends of the long wire. Now put even a tiny load on and now you have voltage drop. The larger the load, the larger the voltage drop in the wires, because the current is higher.

Again obvious you don't understand the basics. Volts are not "lost" If they are and you're so smart, give us the formula for the "lost" volts with no current flowing.

Might I suggest a course in electricity 101? Understand how to apply] Ohm's Law to a simple circuit? Kirchoff's Laws?

- posted on May 30, 2016, 2:58 pm

On Sunday, May 29, 2016 at 10:23:36 PM UTC-4, Diesel wrote:

The heating element looks like a resistor. There is no magical making up. Take a simple circuit with a voltage source V and a resistor of value R1. The current is V/R. Now add an additional resistor in series, R2 that represents the resistance of the wire in our system. The current is now V/(R1+R2), which is less than the current of V/R1.

Like Gfre said, if you look at a heating element, if it's rated for two voltages, the power output is stated lower at the lower voltage.

IDK what parameters they used. I worked out the actual numbers, it's very simple and the voltage drop due to the resistance in the wire is only 2.7%, at 20A, for 100 ft of #12. Why no reply to the actual math? Here it is again:

The resistance of 12 gauge wire is .0016 ohms per foot. 100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V . There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V drop from 240V is just 2.7%, not 5.6%.

No idea what you're even doing here, no loads? yet 1500?

Again, the resistance of #12 is .0016 per ft. The math has been presented above, the voltage drop at the full 20A is just is just 6.4V, which at 240V is just 2.7%. Even using your higher number, which is probably coming from using some high temperature, you still get a voltage drop of only 3.2%. There is no 125 ft, he said the run is under 100 ft.

And you say this is with no load present yet. That's incorrect too. This is with the MAX load the circuit is rated for, the full 20A. With no load, there is no voltage drop period.

Then use #8, #6, etc. Same argument can be made there. It does nothing in terms of safety or proper operation of the heater.

The closer I can get to their expected input

You're really confused here. The heater does not magically adjust to draw more amps. Take this to an extreme. If I put just 24V across those heaters, following your logic instead of drawing 14A at 240V they would draw 10 x 14A = 140A? It doesn't work that way.

BS. If that was the case, no electrical inspector would approve it. You have one, Gfre, telling you that it does meet code and citing NEC to back it up.

More BS.

at near full load over a period of time is that the connection

And even using those numbers, the voltage drop at the full 20A circuit rating is just 3.2%. At the actual load, 14.6 amps, it's just 2.3%, so again there is no problem, no hot wires.

Show us where Ugly's says it's a violation of NEC, unsafe, etc to wire the circuit on a #12.

I haven't seen any 240volt baseboard heater wired with a 12/2

That depends entirely on what the rated capacity of the heaters is. And if you're that familiar with heating elements you should know that there ones that are rated for either 240V or 120V and that at the lower voltage they put out LESS power, they don't magically adjust, pull more current, to put out the same power they do at 240V.

I think the last time I actually observed that

Which again implies that this is destined to fail, when of course it's not.

The heating element looks like a resistor. There is no magical making up. Take a simple circuit with a voltage source V and a resistor of value R1. The current is V/R. Now add an additional resistor in series, R2 that represents the resistance of the wire in our system. The current is now V/(R1+R2), which is less than the current of V/R1.

Like Gfre said, if you look at a heating element, if it's rated for two voltages, the power output is stated lower at the lower voltage.

IDK what parameters they used. I worked out the actual numbers, it's very simple and the voltage drop due to the resistance in the wire is only 2.7%, at 20A, for 100 ft of #12. Why no reply to the actual math? Here it is again:

The resistance of 12 gauge wire is .0016 ohms per foot. 100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V . There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V drop from 240V is just 2.7%, not 5.6%.

No idea what you're even doing here, no loads? yet 1500?

Again, the resistance of #12 is .0016 per ft. The math has been presented above, the voltage drop at the full 20A is just is just 6.4V, which at 240V is just 2.7%. Even using your higher number, which is probably coming from using some high temperature, you still get a voltage drop of only 3.2%. There is no 125 ft, he said the run is under 100 ft.

And you say this is with no load present yet. That's incorrect too. This is with the MAX load the circuit is rated for, the full 20A. With no load, there is no voltage drop period.

Then use #8, #6, etc. Same argument can be made there. It does nothing in terms of safety or proper operation of the heater.

The closer I can get to their expected input

You're really confused here. The heater does not magically adjust to draw more amps. Take this to an extreme. If I put just 24V across those heaters, following your logic instead of drawing 14A at 240V they would draw 10 x 14A = 140A? It doesn't work that way.

BS. If that was the case, no electrical inspector would approve it. You have one, Gfre, telling you that it does meet code and citing NEC to back it up.

More BS.

at near full load over a period of time is that the connection

And even using those numbers, the voltage drop at the full 20A circuit rating is just 3.2%. At the actual load, 14.6 amps, it's just 2.3%, so again there is no problem, no hot wires.

Show us where Ugly's says it's a violation of NEC, unsafe, etc to wire the circuit on a #12.

I haven't seen any 240volt baseboard heater wired with a 12/2

That depends entirely on what the rated capacity of the heaters is. And if you're that familiar with heating elements you should know that there ones that are rated for either 240V or 120V and that at the lower voltage they put out LESS power, they don't magically adjust, pull more current, to put out the same power they do at 240V.

I think the last time I actually observed that

Which again implies that this is destined to fail, when of course it's not.

- posted on May 30, 2016, 4:14 am

snipped-for-privacy@aol.com

I sincerely do apologize for having missed that in my original reply to you. With that said, I'd still have opted for 10/2 to feed them, and come off the feed with a 12/2 for each heater. This reduces heat creation along the wire, extends the life of the chosen wire, ensures the greatest possible amount of power is available to each heater on the circuit.

Although power loss will still occur at the last one, the voltage drop won't be as bad as it would have been with a 12/2 run from the panel; and the feed wire won't run as hot supplying power to an individual heater or all of them at the same time. A bit more costly, but, a decent enough tradeoff imo, that I would personally have used 10/2.

As I said though, at the end of the day, it all looks great from here.

--

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Hmmm. I most certainly don't understand how I can access a copy of a

I sincerely do apologize for having missed that in my original reply to you. With that said, I'd still have opted for 10/2 to feed them, and come off the feed with a 12/2 for each heater. This reduces heat creation along the wire, extends the life of the chosen wire, ensures the greatest possible amount of power is available to each heater on the circuit.

Although power loss will still occur at the last one, the voltage drop won't be as bad as it would have been with a 12/2 run from the panel; and the feed wire won't run as hot supplying power to an individual heater or all of them at the same time. A bit more costly, but, a decent enough tradeoff imo, that I would personally have used 10/2.

As I said though, at the end of the day, it all looks great from here.

MID: <nb7u27$crn$ snipped-for-privacy@boaterdave.dont-email.me>

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- posted on May 30, 2016, 3:51 pm

On Monday, May 30, 2016 at 12:14:44 AM UTC-4, Diesel wrote:

Sadly, that isn't all that you missed and misapplied.

Sadly, that isn't all that you missed and misapplied.

- posted on May 30, 2016, 9:32 pm

Why? Why not just do the sums and work out what's actually needed instead of believing some paper pushing fuckwit?

--

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