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No, we're not, because T is***not*** infinite. T can be ***any*** ***finite*** value. The one thing it
***cannot*** be is infinite precisely because 0 * infinite is undefined -- but the left side of this
equation, D, is NOT undefined. It is very much defined, and it is EQUAL TO ZERO.

NO WE ARE NOT! Where in the equation D = RT do you see any division anywhere? This is***multiplication***. And as I have already noted, transforming D = RT into D / T = R is valid
only for NON-ZERO values of T.

Complete nonsense. D = RT. D = 0, R = 0, substitute any finite value for T and the equation is true.

No, you still don't have it. C isn't a variable in that equation. C is a constant. If M = 0, then E 0 and C doesn't matter.

The only thing you got right here is that you're certainly wasting a lot of time.

Doug Miller explained :

Which is why Ohm's Law is a formula not just an equation.

It looks just like a letter in an equation not a constant in a formula. As a formula the relationships between the terms become important just like they do in Ohm's Law.

Yeah, I guess it's time to stop now.

It's been fun.

And the difference is --?

I suppose it does -- to you. And to everyone else who doesn't understand it.

Whereas I would not say the ball has an absolute velocity of zero, since it's still moving several hundred miles/hour along several vectors (daily rotation around the earths axis, yearly rotation around the suns axis, longer-term rotation around the center of the galaxy, et cetera, et alia).

If you're refering to relative velocity, then you need to give a referent.

#### Site Timeline

- posted on May 31, 2016, 10:15 pm

No, we're not, because T is

NO WE ARE NOT! Where in the equation D = RT do you see any division anywhere? This is

Complete nonsense. D = RT. D = 0, R = 0, substitute any finite value for T and the equation is true.

No, you still don't have it. C isn't a variable in that equation. C is a constant. If M = 0, then E 0 and C doesn't matter.

The only thing you got right here is that you're certainly wasting a lot of time.

- posted on May 31, 2016, 11:14 pm

Which is why Ohm's Law is a formula not just an equation.

It looks just like a letter in an equation not a constant in a formula. As a formula the relationships between the terms become important just like they do in Ohm's Law.

Yeah, I guess it's time to stop now.

It's been fun.

- posted on June 1, 2016, 12:00 am

And the difference is --?

I suppose it does -- to you. And to everyone else who doesn't understand it.

- posted on June 1, 2016, 4:59 am

On Tuesday, May 31, 2016 at 6:08:02 PM UTC-4, FromTheRafters wrote:

Call it formula or equation, you're still wrong, period. You must have taken quibbling 101 instead of algebra 101.

Call it formula or equation, you're still wrong, period. You must have taken quibbling 101 instead of algebra 101.

- posted on May 31, 2016, 6:46 pm

On Tuesday, May 31, 2016 at 1:12:24 PM UTC-4, Doug Miller wrote:

Thanks Doug. One more step in stopping the insanity! We have a real winner here. He doesn't understand basic math and claims that if you can manipulate any equation so that a zero could possibly be in the denominator, then the whole equation is invalid. That makes pretty much all equations invalid.

He also fails to graph that when an equation produces a result of zero, it's often the most interesting, or defining case. For example, we can apply Newton's Law to give the velocity of a ball shot up in the air. At the peak, the equation give a value of Zero. To you and I, that's a valid answer, it means the velocity is zero, the ball isn't moving, etc. To him, I guess it's "undefined".

Thanks Doug. One more step in stopping the insanity! We have a real winner here. He doesn't understand basic math and claims that if you can manipulate any equation so that a zero could possibly be in the denominator, then the whole equation is invalid. That makes pretty much all equations invalid.

He also fails to graph that when an equation produces a result of zero, it's often the most interesting, or defining case. For example, we can apply Newton's Law to give the velocity of a ball shot up in the air. At the peak, the equation give a value of Zero. To you and I, that's a valid answer, it means the velocity is zero, the ball isn't moving, etc. To him, I guess it's "undefined".

- posted on May 31, 2016, 8:30 pm

Whereas I would not say the ball has an absolute velocity of zero, since it's still moving several hundred miles/hour along several vectors (daily rotation around the earths axis, yearly rotation around the suns axis, longer-term rotation around the center of the galaxy, et cetera, et alia).

If you're refering to relative velocity, then you need to give a referent.

- posted on May 31, 2016, 8:41 pm

On Tuesday, May 31, 2016 at 4:30:29 PM UTC-4, Scott Lurndal wrote:

Oh, please, stop with the nonsense. Apparently you never took high school physics either. That type of question is a typical one found on physics tests covering Newton's Laws. Cannon gets fired, ball goes up in the air, object gets dropped. Calculate the velocity, the height, etc. It's on the physics SATs. They don't preface the whole damn thing with a page of disclaimers about not including the motion of the earth, the universe, realtivity considerations, etc. Did you go to school with Rafters?

Oh, please, stop with the nonsense. Apparently you never took high school physics either. That type of question is a typical one found on physics tests covering Newton's Laws. Cannon gets fired, ball goes up in the air, object gets dropped. Calculate the velocity, the height, etc. It's on the physics SATs. They don't preface the whole damn thing with a page of disclaimers about not including the motion of the earth, the universe, realtivity considerations, etc. Did you go to school with Rafters?

- posted on May 30, 2016, 8:57 pm

On 05/30/2016 01:18 PM, FromTheRafters wrote:

[snip]

With any 2 nonzero numbers, A/B is the reciprocal of B/A. In other words, (A/B)*(B/A)=1. If you make A=0, then A/B=0. B/A should then give the reciprocal of 0 (as in, what do you multiply 0 by to get 1). How about infinity?

Still doesn't explain 0/0.

If you don't like this, please don't read it :-)

[snip]

With any 2 nonzero numbers, A/B is the reciprocal of B/A. In other words, (A/B)*(B/A)=1. If you make A=0, then A/B=0. B/A should then give the reciprocal of 0 (as in, what do you multiply 0 by to get 1). How about infinity?

Still doesn't explain 0/0.

If you don't like this, please don't read it :-)

--

Mark Lloyd

http://notstupid.us/

Mark Lloyd

http://notstupid.us/

Click to see the full signature.

- posted on May 30, 2016, 9:19 pm

Mark Lloyd laid this down on his screen :

Maybe you could multiply it by itself zero times. Anything to the zeroeth power equals one.

Okay, so multiplicative groups kick out the zero, it was worth a try.

Maybe you could multiply it by itself zero times. Anything to the zeroeth power equals one.

Okay, so multiplicative groups kick out the zero, it was worth a try.

- posted on May 30, 2016, 9:13 pm

On Mon, 30 May 2016 14:18:28 -0400, FromTheRafters

- posted on May 30, 2016, 9:28 pm

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

Says you. Do you divide by zero for a living?

Says you. Do you divide by zero for a living?

- posted on May 30, 2016, 10:17 pm

On 05/30/2016 04:28 PM, FromTheRafters wrote:

[snip]

I seem to remember n/0 = INF (for nonzero values of n). 0/0 = NaN,

I'm not so sure about that last one, I've seen 0, 1, and infinity. All at the same time. It's a fuzzy number. Very fuzzy.

[snip]

I seem to remember n/0 = INF (for nonzero values of n). 0/0 = NaN,

I'm not so sure about that last one, I've seen 0, 1, and infinity. All at the same time. It's a fuzzy number. Very fuzzy.

- posted on May 30, 2016, 11:55 pm

notX has brought this to us :

You might be correct in that particular context.

https://en.wikipedia.org/wiki/Division_by_zero#Computer_arithmetic

If you are happy with the 'voltage drop' being somewhere between zero and a very large integer (or infinity) then that's a plus for you. ;)

Still, it can't really be a 'voltage drop' without energy being dissipated by the device under consideration.

You might be correct in that particular context.

https://en.wikipedia.org/wiki/Division_by_zero#Computer_arithmetic

If you are happy with the 'voltage drop' being somewhere between zero and a very large integer (or infinity) then that's a plus for you. ;)

Still, it can't really be a 'voltage drop' without energy being dissipated by the device under consideration.

- posted on May 31, 2016, 3:37 am

On Mon, 30 May 2016 19:55:22 -0400, FromTheRafters

acroos the circuit is the supply voltage. Since the voltage drop across the conductors is zero, the voltage drop across the "open" is supply voltage. (the voltage drop MUST equal the impressed voltage) This is true with zero amps current flow. Because there is zero amps current you can not solve for the resistance of the conductor because (supply voltage assumed to be 12) 12/0 is indefineableand the reistance of the open circuit is infinite (in a perfect world)

acroos the circuit is the supply voltage. Since the voltage drop across the conductors is zero, the voltage drop across the "open" is supply voltage. (the voltage drop MUST equal the impressed voltage) This is true with zero amps current flow. Because there is zero amps current you can not solve for the resistance of the conductor because (supply voltage assumed to be 12) 12/0 is indefineableand the reistance of the open circuit is infinite (in a perfect world)

- posted on May 31, 2016, 10:07 am

on 5/30/2016, snipped-for-privacy@snyder.on.ca supposed :

Why call it a voltage drop when you have previously stated several times(correctly) that it requires a current to make a 'voltage drop'?

Why call it a voltage drop when you have previously stated several times(correctly) that it requires a current to make a 'voltage drop'?

- posted on May 31, 2016, 3:06 pm

On Tuesday, May 31, 2016 at 6:07:11 AM UTC-4, FromTheRafters wrote:

Because with no current flowing, the voltage drop, what we're talking about is zero. Did you try plotting a graph for Ohm's Law, like I suggested? It's a straight line through the origin. According to you, for some unknown reason, the origin, where I= 0, V = 0, does not exist, only the rest of the line does.

Next!

Because with no current flowing, the voltage drop, what we're talking about is zero. Did you try plotting a graph for Ohm's Law, like I suggested? It's a straight line through the origin. According to you, for some unknown reason, the origin, where I= 0, V = 0, does not exist, only the rest of the line does.

Next!

- posted on May 31, 2016, 3:43 pm

On Tuesday, May 31, 2016 at 6:07:11 AM UTC-4, FromTheRafters wrote:

Obviously you never took a course in algebra or physics. We still call it a voltage drop of zero for precisely the same reason we call any other property by it's name with a value of zero. Apply Newton's Law to a ball thrown straight up in the air at a speed of 20M/sec. What is the velocity along the way? We can write the equation, calculate the velocity at every point. At the peak, the equation gives a value of ZERO. That value of zero, just like a voltage drop of zero, has meaning. High school test question:

What is the value of the velocity at the max height?

Answer of the rest of the word: Zero

Your answer: There is no meaning to velocity.

Obviously you never took a course in algebra or physics. We still call it a voltage drop of zero for precisely the same reason we call any other property by it's name with a value of zero. Apply Newton's Law to a ball thrown straight up in the air at a speed of 20M/sec. What is the velocity along the way? We can write the equation, calculate the velocity at every point. At the peak, the equation gives a value of ZERO. That value of zero, just like a voltage drop of zero, has meaning. High school test question:

What is the value of the velocity at the max height?

Answer of the rest of the word: Zero

Your answer: There is no meaning to velocity.

- posted on May 31, 2016, 8:40 pm

On Tue, 31 May 2016 06:07:03 -0400, FromTheRafters

drop. Doesn't matter what the theoretical math is.

drop. Doesn't matter what the theoretical math is.

- posted on May 31, 2016, 3:25 am

On Mon, 30 May 2016 17:28:23 -0400, FromTheRafters

measuring the voltage you are introducing a non-zero value to both resistance and current. - Resistance WAY off from zero - approaching (but never COMPLETELY reaching infinite (by the very definition of infinity) meaning current- for all practical reasons being ZERO - but in reality just being infinitesimally small, Do the calculations using a current of 0.000000(100,000,000 zeros)01 amps and the resistance being 0.0000000100,000,000 zeros)01 ohms for an open circuit and everything works.

Sane as with a dead short=0.99999o(1000,000,000 nines)99 ohms because we don't have superconductors. In reality you don't need to go nearly as far ac the impedence of any meter is sigmificantly lower than that - with sensitivity being in the megohms per volt range.on digitals and kilohms per volt on analogs.

measuring the voltage you are introducing a non-zero value to both resistance and current. - Resistance WAY off from zero - approaching (but never COMPLETELY reaching infinite (by the very definition of infinity) meaning current- for all practical reasons being ZERO - but in reality just being infinitesimally small, Do the calculations using a current of 0.000000(100,000,000 zeros)01 amps and the resistance being 0.0000000100,000,000 zeros)01 ohms for an open circuit and everything works.

Sane as with a dead short=0.99999o(1000,000,000 nines)99 ohms because we don't have superconductors. In reality you don't need to go nearly as far ac the impedence of any meter is sigmificantly lower than that - with sensitivity being in the megohms per volt range.on digitals and kilohms per volt on analogs.

- posted on May 31, 2016, 10:13 am

It happens that snipped-for-privacy@snyder.on.ca formulated :

Who said anything about measuring? It is well known that the act of detecting a thing affects the thing being detected. We were talking about the existence of 'voltage drop' when there is zero (by definition) current.

[snipped the what if almost zero scenarios]

Who said anything about measuring? It is well known that the act of detecting a thing affects the thing being detected. We were talking about the existence of 'voltage drop' when there is zero (by definition) current.

[snipped the what if almost zero scenarios]

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