The bistander thread

[Turion]

[bistander]

$

So you believe the ConFlow works as claimed? Seems like you should put some $ on it.

[Turion]

ConFlow

I know it is possible to get out significantly MORE than you put in. So I know that what ConFlow is doing is possible.

As for investing in this technology, I already am.

bi, right now YOU know everything you need to know to have a working free energy device that does not violate a SINGLE one of the laws of thermodynamics sitting on your bench. You just haven’t put it all together. Some day you will, or you will see it operating somewhere and you will be kicking yourself. I know you will say that getting out more than you put in violates their laws, but ONLY when applied to closed systems, not OPEN ones. Think of it this way. We are creating an energy “sail” that captures energy rather than wind.

[Quantum_well]

Answer.

I think you'll find the answers in the name

ConFlow..,.......Con trick!

[Turion]

Closed Minds

Just because YOU don't know how to do it doesn't mean it can't be done. You guys are stuck in a box. It is so unfortunate that those of us outside the box have to listen to all the noise you make because it is evident you will never escape.

[bistander]

Box

Just because YOU don't know how to do it doesn't mean it can't be done. You guys are stuck in a box. It is so unfortunate that those of us outside the box have to listen to all the noise you make because it is evident you will never escape.

Hi Turion,

What makes you think you're outside the box? You don't even know what the box is. Your last sentence in your prior post typifies that.

Sail on,

bi

[Turion]

Lol

I intend to. Thanks!
Look up the word metaphor. You may have heard of it.
(You know better because I KNOW you know what a metaphor is. You just like to argue)

[bistander]

Metaphor

I intend to. Thanks!
Look up the word metaphor. You may have heard of it.
(You know better because I KNOW you know what a metaphor is. You just like to argue)

It's easy to read into your metaphor that you infer that the real sail captures wind and not energy.

And as usual, you don't answer the question. I don't like to argue, I like to see a guy stand behind what he says by providing proof of his claim.

bi

[Turion]

Rough seas

No time to talk. Too busy sailing. Avast matey, you landlubbers need to stay on shore. Ye be too afraid of gettin yer feet wet! Har har har

[bistander]

Jettis example coil

Hi Dave,

I would have preferred to analyze just a coil, not a motor, but I'll give it a try.

You have a 24V motor which runs 1000RPM drawing 210 mA. I presume this is running at no-load and that load doesn't change. Initially you use a single strand winding. When you increase that to 2 strands in parallel, the current doesn't change however the RPM increase ~3%. Fair statement?

Going from 1 strand to 2 halves the coil resistance. You have doubled the utilized copper in the coil. This increase of copper increases efficiency, or in other words, decreases loss (I^2R). This halving of resistance also increases the applied armature voltage due to cutting the resistive drop in half.

As for the observed performance changes, current doesn't change. Motor current is dependent on motor load (torque) which at no-load, as the case here, consists of rotor friction and aero drag. So for a few percent Chang in RPM, current essentially remains same.

The increase in RPM is due to the reduced voltage drop in the coil. The source voltage remains the same but less voltage drop in the coil means a higher voltage applied to the armature. Since this armature voltage must equal the flux times velocity per Faraday, the velocity (RPM) increases.

No surprises.

Regards,

bi

{edit}

Sorry folks... I was going from memory just got home and checked my lab notes, on the above quote I was wrong the first strand was running at 170mA (907 rpm) I then connected the second strand the current increased to 200mA, the third 210mA, 4th 210mA, 5th 220mA... 10th 210mA (1157rpm) I can post the lab note this afternoon FYI.

Dave Wing

Dave just update data from his notes. This is still consistent with my description. The RPM increase is larger that first stated. This increases the developed torque to spin at no-load and hence the proportionate increase in current.

[Turion]

Question

bi,
When a motor is connected to a battery, how can the amps measured on the positive line be the same as those measured on the negative line if the load is consuming power?

An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second.

Current is a count of the number of electrons flowing through a circuit. One amp is the amount of current produced by a force of one volt acting through the resistance of one ohm.]

Just wondering what your thoughts are on this matter.

[bistander]

Electric circuit basics

Pop goes the weasel, and bi speaks! I knew he would. running his mouth is all he ever does. No building, no researching. Just blathering.

Yes bi, I HAVE read the physics definition of potential. Unfortunately for YOU we are dealing with the electrical definition of potential, which is simply the charge in an electrical circuit. You can't even get the 3 battery system to work and you're going to lecture ME about potential? I wonder if you realize how often the few of us who have working systems talk about you and LAUGH out loud at how moronic some of your comments are and how LITTLE you actually know?
...

bi,
When a motor is connected to a battery, how can the amps measured on the positive line be the same as those measured on the negative line if the load is consuming power?

An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor. One ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second.

Current is a count of the number of electrons flowing through a circuit. One amp is the amount of current produced by a force of one volt acting through the resistance of one ohm.]

Just wondering what your thoughts are on this matter.

bi, ... how can the amps measured on the positive line be the same as those measured on the negative line if the load is consuming power? ...

Hello Turion,

You certainly have some audacity to ask me (what seems like) a serious question after all the insults and ridicule which you direct my way. No doubt I'm a fool to give a serious answer, but wtf. How can current be the same in and out of the load when that load is consuming power? Answer is potential.

I think I mentioned before; you really would benefit from a basic course in electricity which includes circuits.

Regards,

bi

[jettis]

Hi Dave,

I would have preferred to analyze just a coil, not a motor, but I'll give it a try.

You have a 24V motor which runs 1000RPM drawing 210 mA. I presume this is running at no-load and that load doesn't change. Initially you use a single strand winding. When you increase that to 2 strands in parallel, the current doesn't change however the RPM increase ~3%. Fair statement?

Going from 1 strand to 2 halves the coil resistance. You have doubled the utilized copper in the coil. This increase of copper increases efficiency, or in other words, decreases loss (I^2R). This halving of resistance also increases the applied armature voltage due to cutting the resistive drop in half.

As for the observed performance changes, current doesn't change. Motor current is dependent on motor load (torque) which at no-load, as the case here, consists of rotor friction and aero drag. So for a few percent Chang in RPM, current essentially remains same.

The increase in RPM is due to the reduced voltage drop in the coil. The source voltage remains the same but less voltage drop in the coil means a higher voltage applied to the armature. Since this armature voltage must equal the flux times velocity per Faraday, the velocity (RPM) increases.

No surprises.

Regards,

bi

{edit}


Dave just update data from his notes. This is still consistent with my description. The RPM increase is larger that first stated. This increases the developed torque to spin at no-load and hence the proportionate increase in current.

Here are my lab notes, the second image is when I put another coil in parallel with the first, using the same circuit. Rpm increased to 1460 from 1159 and current draw went down to 190mA from 220mA.

[bistander]

Here are my lab notes, the second image is when I put another coil in parallel with the first, using the same circuit. Rpm increased to 1460 from 1159 and current draw went down to 190mA from 220mA.

Hi Dave,

I'd help if I could see the apparatus. Also, what is it that you think is "magical"?

Regards,

bi

[Turion]