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Renewable Energy Discussion on various alternative energy, renewable energy, & free energy technologies. Also any discussion about the environment, global warming, and other related topics are welcome here.

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  #301  
Old 10-28-2019, 07:21 PM
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ConFlow

LOL. Can’t work anyway according to YOU. More out than in is claimed. You people have NO IDEA what is possible, and no laws are broken. A few are bent a little because things are done in a different way, but none broken.
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  #302  
Old 10-28-2019, 07:35 PM
bistander bistander is offline
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$

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LOL. Can’t work anyway according to YOU. More out than in is claimed. You people have NO IDEA what is possible, and no laws are broken. A few are bent a little because things are done in a different way, but none broken.
So you believe the ConFlow works as claimed? Seems like you should put some $ on it.

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ConFlow Power finalized its listing on the early adopter’s board within the ILOCX ecosystem priced at $20.00 per license that will bring in a total of $2,000,000 in pre-sales for the ILO.
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  #303  
Old 10-29-2019, 02:21 AM
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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.
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Last edited by Turion; 10-29-2019 at 02:57 AM.
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  #304  
Old 10-29-2019, 08:19 AM
Quantum_well Quantum_well is offline
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Answer.

I think you'll find the answers in the name

ConFlow..,.......Con trick!
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  #305  
Old 10-29-2019, 05:38 PM
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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.
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  #306  
Old 10-29-2019, 07:03 PM
bistander bistander is offline
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Box

Quote:
Originally Posted by Turion View Post
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.
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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.

Quote:
We are creating an energy “sail” that captures energy rather than wind.
Sail on,

bi
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  #307  
Old 10-29-2019, 09:14 PM
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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)
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Last edited by Turion; 10-29-2019 at 09:20 PM.
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  #308  
Old 10-30-2019, 02:54 AM
bistander bistander is offline
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Metaphor

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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)
Quote:
We are creating an energy “sail” that captures energy rather than wind.
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
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  #309  
Old 10-30-2019, 06:13 AM
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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
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  #310  
Old 12-03-2019, 01:48 PM
bistander bistander is offline
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Jettis example coil

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Originally Posted by jettis View Post
On my bench I have a SSG single coil machine, the coil consists of 68 strands of 26AWG at a hundred feet. Six of these 26AWG strands are connected in series for a total of 600’ in length. I made 5 more 600’ strands using this pattern. This arrangement has given me a total of 10 strands of multifilar wire at 600’ long and eight 100’ leftover windings extra, of which one of these is then used for a trigger.

Anyone can build this it’s easy and cheap to build.

On the bench the 24 volt SSG 4 pole junior machine is first run with one 600’ length, under this arrangement it will run at 210mA at 1000’ rpm on one strand of 600’ of mutilfilar wire, when tuned to a single pulse. As each successive multifilar strand is added in parallel to the SSG there is a increase of anywhere from 20 to 40 RPM.

Big deal right? You would expect a small current increase with the addition of each strand would you not?

What would you say if there was a no current increase (current stayed at 210mA) with the addition of each 600’ multifilar strand in parallel and the only increase was in magnetic field strength ( which translated into around a 300rpm increase) when additional strands were added?

Dave Wing
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}
Quote:
Originally Posted by jettis View Post
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.
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  #311  
Old 12-04-2019, 10:31 PM
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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.
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Last edited by Turion; 12-04-2019 at 11:18 PM.
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  #312  
Old 12-05-2019, 02:09 AM
bistander bistander is offline
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Electric circuit basics

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Originally Posted by Turion View Post
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?
...
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Bi,
Believing textbooks are always right is why you will never find free energy.

As to batteries....Just because you have charged a battery doesn’t mean you understand how they work or what the energy in them is capable of doing. It is very clear that you do not. Believe me, I take GREAT pleasure in knowing that you have NO clue. ...

Quote:
Originally Posted by Turion View Post
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.
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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
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  #313  
Old 12-05-2019, 02:31 AM
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Originally Posted by bistander View Post
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.
Attached Images
File Type: jpg 40F044E6-33E2-45C0-89F9-FCE1EF449F19.jpg (129.4 KB, 10 views)
File Type: jpg 074D5B13-6138-4AE5-95FE-03D3C189F774.jpg (157.6 KB, 10 views)
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  #314  
Old 12-05-2019, 03:36 AM
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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
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  #315  
Old 12-05-2019, 05:17 AM
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Another question

Yeah, audacity is my middle name. As to learning about electricity, I’m pretty sure I know a few things you don’t, despite the fact that I would definitely benefit from some courses in basic electricity. Maybe. Possibly. But then again it’s probably not necessary. My research into free energy is done. I won’t be building any new electrical circuits. I know what I need to know. I’m right where I want to be.

Despite what you believe, my purpose here has ALWAYS been to help others as much as I can.

But back to the subject at hand. So you don’t believe that the amp reading is based on the resistance in the circuit, and because of resistance the amp reading is the same at any point in the circuit?

Can we agree that if power is being consumed by the load, what we will or SHOULD see is a voltage drop? If energy is being consumed by the load, SOMETHING has to go down, correct? If it isn’t the amp reading, which is the same on both sides of the motor, should you not then expect to see a voltage drop from one side of the motor to the other? And how would you check to see if that is actually happening?
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  #316  
Old 12-05-2019, 07:51 AM
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Originally Posted by bistander View Post
Hi Dave,

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

Regards,

bi
The machine is nothing special it is the 4 pole Junior kit Rick Friedrich sold when he was with John. I use the normal 3-1/2 inch Pittsfield spools, as supplied in the kit. I will show some footage at some point, when I have time.

What typically happens to the amp draw and the rpm of the rotor when we add another salient pole to a motor, in parallel?

Dave Wing
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File Type: jpg FDC4AC4A-23C5-410B-A6C9-FAD8576B5EF3.jpg (193.4 KB, 19 views)
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  #317  
Old 12-05-2019, 01:43 PM
bistander bistander is offline
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Quote:
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Yeah, audacity is my middle name. As to learning about electricity, I’m pretty sure I know a few things you don’t, despite the fact that I would definitely benefit from some courses in basic electricity. Maybe. Possibly. But then again it’s probably not necessary. My research into free energy is done. I won’t be building any new electrical circuits. I know what I need to know. I’m right where I want to be.

Despite what you believe, my purpose here has ALWAYS been to help others as much as I can.

But back to the subject at hand. So you don’t believe that the amp reading is based on the resistance in the circuit, and because of resistance the amp reading is the same at any point in the circuit?

Can we agree that if power is being consumed by the load, what we will or SHOULD see is a voltage drop? If energy is being consumed by the load, SOMETHING has to go down, correct? If it isn’t the amp reading, which is the same on both sides of the motor, should you not then expect to see a voltage drop from one side of the motor to the other? And how would you check to see if that is actually happening?
You obviously know the answers. Get to the point.
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  #318  
Old 12-05-2019, 04:01 PM
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Point

My point is, in the example I gave of a motor connected to a battery, how do you prove there is a voltage drop across the load and that the motor is CONSUMING the energy? Or are you ASSUMING the motor is consuming energy because when measuring the voltage across the battery (which is also measuring across the motor) the measurement is going down?

This is a SERIOUS question and a serious discussion.
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  #319  
Old 12-05-2019, 04:36 PM
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Seriously

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My point is, in the example I gave of a motor connected to a battery, how do you prove there is a voltage drop across the load and that the motor is CONSUMING the energy? Or are you ASSUMING the motor is consuming energy because when measuring the voltage across the battery (which is also measuring across the motor) the measurement is going down?

This is a SERIOUS question and a serious discussion.
O.K. Turion,

First off. I don't like "consuming energy". A motor, in operation as a motor, meaning current flowing through it and a potential difference (voltage) across it, converts electrical power into mechanical power and heat.

Where is this assumption to which you refer? What do you mean "the measurement is going down"?

Regards,

bi
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  #320  
Old 12-05-2019, 05:30 PM
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Definitions again?

Ok, so a motor does not “consume energy” it converts it to mechanical energy. How do you KNOW this is happening? If the amps remain the same there has to be a loss in voltage if some of that electrical energy is converted doesn’t there? How do you measure with your volt meter how much the voltage has been reduced?
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  #321  
Old 12-05-2019, 05:39 PM
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Measuring voltage

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Ok, so a motor does not “consume energy” it converts it to mechanical energy. How do you KNOW this is happening? If the amps remain the same there has to be a loss in voltage if some of that electrical energy is converted doesn’t there? How do you measure with your volt meter how much the voltage has been reduced?
Read the directions which came with your voltmeter or multimeter. Did you try putting one probe on each motor terminal?
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  #322  
Old 12-05-2019, 05:57 PM
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Meter

So you’re telling me that connecting my volt meter across the motor will tell me how much the voltage has been reduced? And knowing this there is a formula for how much energy has been converted to mechanical energy? Or do you base your assumptions on what is now “missing” from the battery? I am trying to find out WHAT I your experience and professional training tells you, or how we can ACCURATELY measure how much of the energy, voltage, amperage, whatever you choose to call it, went into the motor from the battery and what % of that was converted to mechanical energy and how you KNOW that. I’m NOT trying to be an ass. I’m trying to have a real conversation for once about something that is really important.
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  #323  
Old 12-05-2019, 06:45 PM
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Motor power

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So you’re telling me that connecting my volt meter across the motor will tell me how much the voltage has been reduced? And knowing this there is a formula for how much energy has been converted to mechanical energy? Or do you base your assumptions on what is now “missing” from the battery? I am trying to find out WHAT I your experience and professional training tells you, or how we can ACCURATELY measure how much of the energy, voltage, amperage, whatever you choose to call it, went into the motor from the battery and what % of that was converted to mechanical energy and how you KNOW that. I’m NOT trying to be an ass. I’m trying to have a real conversation for once about something that is really important.
O.K. Turion,

Let's talk about a simple DC motor. And a constant voltage power supply, or very large battery where its voltage doesn't change over the duration of interest. Let's call it 12V constant.

Now hook up the battery to the motor using 2 wires. The battery still has 12V measured across its terminals, + & -, the motor has 12V across its terminals and for instance, there is a current of 1 ampere in each wire. Assumption is that wires are large enough and short enough that wire resistance is negligible.

So, there is 12 watts drawn from the battery and 12 watts delivered to the motor, 12V * 1A in both cases. The motor has an input power of 12W. If there is nothing connected to the shaft, it is running at no-load and output power is zero. All 12W of input power is converted to heat. Motor efficiency is power out / power in = 0 / 12W = 0%.

If there is a load put on this motor, that is something attached to the motor shaft opposing rotation, like a fan blade, then the current will increase, let's say to 3A. Now the input power to the motor is 36W and power out of the battery is 36W, 12V * 3A. To find the actual outpower of the motor, you would need a dynamometer to measure it, or a motor performance characteristic curve to look it up. By either method, let's say shaft output power, torque * rotational speed, is 20W. Now the motor is operating at 20W / 36W = 55.6% efficiency with 16W being wasted (converted to heat).

To measure the actual mechanical power out of a motor is difficult. I referred to a dynamometer. Not easy to come by. RPM is the easy part, but measuring torque at speed is difficult. Torque sensors are costly. Simple and crude method is a prony brake. With a simple brushed PM motor like those scooter motors, testing and drawing a characteristic performance curve wouldn't be too complicated.

Hope that answers your question.

bi
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  #324  
Old 12-05-2019, 07:37 PM
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Loss of voltage

My original statement was that if amperage was always the same at all points in the circuit, you would only know the motor was expending or “converting” energy if the voltage went down. In your example the battery is big enough that voltage is constant, so your measurement of what energy was “converted” is based on two things:
1. Amps x volts= watts (over time)
3. The efficiency of the motor based on a test that would have to be run.

Is that a fair assessment?


Is it fair to say that you are basing your estimate of how much energy was converted based on what came out if the battery and how efficiently the motor ”converted” that energy (based on the test you talked about) into mechanical energy?
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Old 12-05-2019, 08:43 PM
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Test............ Xxxxx

Edit test.

Still can't post.

Another test....

And another test.
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  #326  
Old Yesterday, 03:16 AM
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Test post

Still can't edit or post. This is just another test.

bi
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  #327  
Old Yesterday, 04:29 AM
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Lightbulb

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Originally Posted by Turion View Post
My original statement was that if amperage was always the same at all points in the circuit, you would only know the motor was expending or “converting” energy if the voltage went down. In your example the battery is big enough that voltage is constant, so your measurement of what energy was “converted” is based on two things:
1. Amps x volts= watts (over time)

3. The efficiency of the motor based on a test that would have to be run.
If you know the motor efficiency, you can determine the portion of that energy converted to mechanical energy and that converted to heat.

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Is that a fair assessment?


Is it fair to say that you are basing your estimate of how much energy was converted based on what came out if the battery and how efficiently the motor ”converted” that energy (based on the test you talked about) into mechanical energy?
What estimate? I outlined how to measure it.

It appears that you are attempting to read something into my reply or put words in my mouth. What is your motive here?

bi
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Old Yesterday, 05:22 AM
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Is it fair to say that you are basing your estimate of how much energy was converted based on what came out if the battery and how efficiently the motor ”converted” that energy (based on the test you talked about) into mechanical energy?
Not trying to put words in your mouth. Is it fair to say that your COMPUTATION of how much energy was converted is based on what came out if the battery and how efficiently the motor ”converted” that energy (as determined by the test you talked about) into mechanical energy?

In other words, you believe the electrical energy that came out of the battery and went into the motor was basically ALL converted (dependent on the motor’s efficiency rating) into mechanical energy. That all the energy that came out of the battery was either converted to mechanical energy or lost due to the inefficiency of the conversion process. Is that correct?
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Last edited by Turion; Yesterday at 12:00 PM.
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  #329  
Old Yesterday, 09:41 AM
Quantum_well Quantum_well is offline
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Posts: 42
If your motor is not turning anything you have made
a device to heat your shop,even the sound made ends
up as heat.
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  #330  
Old Yesterday, 12:49 PM
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BroMikey BroMikey is offline
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Quote:
Originally Posted by Quantum_well View Post
If your motor is not turning anything you have made
a device to heat your shop,even the sound made ends
up as heat.

I think it is interesting what Dave is saying about current in and voltage
in then current out is same and voltage, motor is suppose to be burning
up THAT amount measured on a meter. So hey just send the current
around and around again, right? Sounds like we just split the positive,
doesn't it?

Of course the voltage drop is not to be ignored.

Where are the thinkers who can address the material presented instead
of side stepping every point? Huh? Where? Just change the subject I know
I know. The motor is hot the motor is cold the motor is pink too.

Answer the questions or forget the whole think. I know go read another
book, right?

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