The bistander thread

[jettis]

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

[bistander]

You obviously know the answers. Get to the point.

[Turion]

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.

[bistander]

Seriously

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

[Turion]

[bistander]

Measuring voltage

Read the directions which came with your voltmeter or multimeter. Did you try putting one probe on each motor terminal?

[Turion]

[bistander]

Motor power

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

[Turion]

[bistander]

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.

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

[Turion]

Not trying to put words in your mouth. Is it fair to say that your COMPUTATION

[Quantum_well]

If your motor is not turning anything you have made
a device to heat your shop,even the sound made ends
up as heat.

[BroMikey]

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?

[Turion]

Restriction

But here is a theory for you, and I hope you will think about it. I sincerely hope you will think about it. I mean I really, really hope you will think about it. This is the closest I can get to pointing you in the absolute right direction, so what you do with it is up to you. You probably won't think about it, because I believe you are a disinformation expert who is here trying to impede the learning process of others, but on the off chance that I am WRONG about that, I am giving you the benefit of the doubt.

Take a fully charged battery and let it set on the shelf in your shop for several months. When you go to use it, it is dead. Why is that? Now you can go to your textbooks and read about chemical reactions in batteries and how the electrical energy is converted to heat in a chemical process or other "theories" about what happened to that charge or “energy”. But they are only theories. And here's mine:

What if energy is like water? It wants to flow and seek its own level. And what if when we charge a battery, all we are doing is forcing all that energy to one side of the battery. Which is why it takes SO much extra energy to charge a battery. You are probably not putting much of a "charge" into it at all. You have to POUND it into submission and move the EXISTING energy around. (You can call this energy ions, or electrons or marbles. I don't care. I'm talking about CONCEPTS here and I don't need a vocabulary lesson) When you put your meter on the battery you are measuring NOT what is in the battery, but the difference between one side and the other. One side is at ZERO volts and the other is at 12 (for example).

[Turion]

continued

The electrolyte is made up of distilled water, which is non conductive, and a chemical solution that is SLIGHTLY conductive. So as the battery sits, that energy moves slowly from the "high" side to the "low" side of the battery through this slightly conductive electrolyte, attempting to equalize, and the reading on the meter, measuring the difference, continues to go down as the battery equalizes. When the battery goes "dead" there is no current flow between the two sides because they are close to equal. And you meter reflects this.

Now, when you put a load on the battery, such as the motor, you provide a path of less resistance for this current flow. Exactly the SAME thing happens as when a battery is left sitting, only much faster. You start with 12 volts difference between the two sides (for example), but as they equalize THROUGH THE LOAD, your meter reads less and less until there is not enough current flow for the motor to run. NO ENERGY from the battery was "CONVERTED" to mechanical energy at all. The motor worked because current flowed through the wires from one side of the battery to the other, and quit working when there was no more current flow because the two sides were equalizing and the flow was so little, the motor would no longer run on it. But the key is that all the energy is still THERE. It is just equally divided between the two sides. I hope you see the two most important points here.
1. NOTHING was "consumed, used up or converted by the load" THIS IS THE ONLY POINT where anything Matt and I have done diverges from accepted electrical theory.
2. The load ran as long as there was current flow from high to low.