Left Hand Rule Applied to a Winding

[jake jot]

Two LEDs connected in antiparallel.As a general approximation, the strength of the magnetic field is proportional to the product of current times turns, called Ampere-Turns. You would need big batteries and larger wire connecting them if you want few turns and many amps.

Are the formulas for transformer windings the same as the winding in motors? I'm googling about the formulas and I read about inductance. Maybe the single conductor won't work because it would just short the batteries? So all these illustrations in youtube won't work at all because they would short?

Also I tried to determine the force of the rotors using the left hand rule. I can point my right hands better. That's when I read the rule for generator is difference. I read this:

"Fleming's two rules are very similar, but they must not be confused. When a current causes motion (like in a d.c. motor), Fleming's left hand rule applies. When motion causes current (like in a generator), Fleming's right hand rule applies. Since these two phenomena are the exact opposite from each other, it makes sense that direction of current would be opposite from each other. If the rules were not opposite then we could easily make electrically-driven perpetual motion machines. "

Can you give illustration how if the rules were not opposite, we could easily make electrically-driven perpetual motion machines? Just to understand the concept of the oppositeness.

 

[jake jot]

Are the formulas for transformer windings the same as the winding in motors? I'm googling about the formulas and I read about inductance. Maybe the single conductor won't work because it would just short the batteries? So all these illustrations in youtube won't work at all because they would short?

View attachment 274510

Also I tried to determine the force of the rotors using the left hand rule. I can point my right hands better. That's when I read the rule for generator is difference. I read this:

"Fleming's two rules are very similar, but they must not be confused. When a current causes motion (like in a d.c. motor), Fleming's left hand rule applies. When motion causes current (like in a generator), Fleming's right hand rule applies. Since these two phenomena are the exact opposite from each other, it makes sense that direction of current would be opposite from each other. If the rules were not opposite then we could easily make electrically-driven perpetual motion machines. "

Can you give illustration how if the rules were not opposite, we could easily make electrically-driven perpetual motion machines? Just to understand the concept of the oppositeness.

To add. Here is some weird results.

I tried to add all my 8 rechargeable batteries (1.2 x 8 = 9.8v) (all newly charged) waiting for the rotor winding to burn (so I can try winding new one for once in my life). But the motor kept running. I measured the amperage in series to the 8 batteries and it reads 1A average. But when I used the hand crank turning the same unit into generator (original design), the output reads only at most 10mA (with the 2 leds removed). And voltage is only 0.5v! I can't even get it above 1v. How do you explain this?

Is it not if the rotor is a certain given speed. It should output the same voltage or current as when you use the battery of same voltage and current to turn the rotor (at same speed)?The wiring diameter of the rotor is about 0.20mm (0.0078 inches). Again this is how it looks like. How do you know the maximum battery voltage it can take? I no longer have extra batteries to test winding burnout voltage.

 

[Merlin3189]

First, your LEDs are diodes. Each allows the current in one direction only. So depending on the polarity of the supply, either red or green lights
.

It was only when I turned the unit sideways that the rotor can rotate (see above video).
Maybe less friction??
I removed the leads less than 3 seconds as I didn't want the winding to burn up. I want to know the following:

1. Can the winding be just a single conductor? .
What's the problem with single conductor?

2. What is the formula for numbers of winding and size? Is there a formula or people just do it by trial and error?

3. I'll only remove the windings (destroyed it) and replace with a single conductor if there is theoretical reason the conductor can work too.

4. This proves generator and motor has relationship.

1 - Yes, maybe
- You have to put a lot of current through it. If you have 100 turns with 10 mA, that's equivalent to 1 turn with 1000 mA. If it took say 100 mA with 100 turns, then you'd need to put 10 A through a single turn.

2 - You'd need to know the strength of the magnetic field in order to calculate the torque. But the force is proportional to the length of the side parallel to the axis and to the aggregate current in the winding, assuming it's all in the same field (With a fixed size magnet, you don't get more force once your coil is as wide as the magnet.)

3 - Usually we prefer more turns of thinner wire. If you halve the CS area of the wire, have double the number of turns and halve the current, then you get the same force. Resistance is double, current is half, so voltage is the same, but power lost to resistance is less, because it's $I^2R$ or $\frac V R$

4 - Abso-blooming-lutely! The motor IS a generator, even when you are using it as a motor. That's where the "back emf" comes from. That's why the voltage determines the speed - when the generator voltage (plus resistance voltage) equals the battery voltage, the battery can't push more current through, so torque can't increase and it can't accelerate. (A bit over simple, but more or less.)Edit: changed layout as it didn't look ok when posted.

 

[jake jot]

First, your LEDs are diodes. Each allows the current in one direction only. So depending on the polarity of the supply, either red or green lights
.View attachment 274565

1 - Yes, maybe
- You have to put a lot of current through it. If you have 100 turns with 10 mA, that's equivalent to 1 turn with 1000 mA. If it took say 100 mA with 100 turns, then you'd need to put 10 A through a single turn.

2 - You'd need to know the strength of the magnetic field in order to calculate the torque. But the force is proportional to the length of the side parallel to the axis and to the aggregate current in the winding, assuming it's all in the same field (With a fixed size magnet, you don't get more force once your coil is as wide as the magnet.)

3 - Usually we prefer more turns of thinner wire. If you halve the CS area of the wire, have double the number of turns and halve the current, then you get the same force. Resistance is double, current is half, so voltage is the same, but power lost to resistance is less, because it's $I^2R$ or $\frac V R$

4 - Abso-blooming-lutely! The motor IS a generator, even when you are using it as a motor. That's where the "back emf" comes from. That's why the voltage determines the speed - when the generator voltage (plus resistance voltage) equals the battery voltage, the battery can't push more current through, so torque can't increase and it can't accelerate. (A bit over simple, but more or less.)Edit: changed layout as it didn't look ok when posted.

Thanks. I made this video and thinking whether to include it in the amazon review of the product. None mentioned it can be turned into motor.



Many customers are children. And maybe for safety. I'd not post it. So don't make the above public.

Here is the mystery. Putting a multimeter in series to the batteries, the current of the above setup is 1A. But using the hand crank, no matter what rotations, it can't output more than 10mA and 0.5volts. If you will notice in the video, the rotations are not that fast. Well. Even if it's fast when put sideways to avoid friction. The handcrank can make it as fast. Is it possible or the case that in a generator. You need more input power to make it match the voltage and current you would use when using it as motor?

 

[Merlin3189]

First - don't worry about children. Using it as a generator they can't get a significant voltage, so no danger.

Putting a multimeter in series to the batteries, the current of the above setup is 1A. But using the hand crank, no matter what rotations, it can't output more than 10mA and 0.5volts.

The generator emf should depend only on speed.
When using it as a motor, the applied voltage has to match the generator emf plus the Ohmic emf caused by the current passing through the windings (and other bits.)
Looking at the 1 A current with 9.8 V battery, the resistance (windings, connections, battery internal resistance) is less than 9.8 Ohm. How much less depends on the motor speed. Maybe your multimeter could measure the resistance at zero speed?

Looking at the open circuit voltage of 0.5 V at max cranking speed, the short circuit current is 10 mA, which implies a resistance of $\frac {0.5}{0.01) = 50 Ohm$ , which includes the winding, contacts and ammeter.
So, a big discrepancy, for which I have to guess at possible causes.
Is it rotating at the same speed when measuring voltage and when measuring current? There is no load other than friction, when measuring o/c voltage. When measuring s/c current, there will be a load torque added, so it may be going slower and the voltage is really less than the 0.5 V max. But 5 times slower ?
Hand cranking probably gives a varying reading, which will be harder to take accurately.
Is any current still going through the dead LEDs? (But I think you've taken them off.)

As far as I can see, 9.8 Ω must be greater than the winding resistance, so something must be wrong with the generator measurement.

 

[jake jot]

First - don't worry about children. Using it as a generator they can't get a significant voltage, so no danger.

The generator emf should depend only on speed.
When using it as a motor, the applied voltage has to match the generator emf plus the Ohmic emf caused by the current passing through the windings (and other bits.)
Looking at the 1 A current with 9.8 V battery, the resistance (windings, connections, battery internal resistance) is less than 9.8 Ohm. How much less depends on the motor speed. Maybe your multimeter could measure the resistance at zero speed?

Looking at the open circuit voltage of 0.5 V at max cranking speed, the short circuit current is 10 mA, which implies a resistance of $\frac {0.5}{0.01) = 50 Ohm$ , which includes the winding, contacts and ammeter.
So, a big discrepancy, for which I have to guess at possible causes.
Is it rotating at the same speed when measuring voltage and when measuring current? There is no load other than friction, when measuring o/c voltage. When measuring s/c current, there will be a load torque added, so it may be going slower and the voltage is really less than the 0.5 V max. But 5 times slower ?

What is meaning of o/c voltage? overclock? or open circuit voltage? or over charge? Also what is the s/c in "s/c current", shut current? step current?

I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms. I put the multimeter in AC instead of DC and tried hand cranking it so fast i don't mind if the rubber band would snap and highest I can get is 1.2 volts and 25mA. If it is set in DC, the voltage highest is 0.7 volts and 20mA. Why does putting multimeter in AC gives higher voltage?

Whereas the voltage (in generator mode) that can even make the rotor rotates by putting it sideways (to avoid friction) should be at least 7 volts dc and it's not rotating as fast. Why couldn't I get it to rotate at 1.2 volts? (i put it sideway to lessen friction from bottom).

What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?

Hand cranking probably gives a varying reading, which will be harder to take accurately.
Is any current still going through the dead LEDs? (But I think you've taken them off.)

As far as I can see, 9.8 Ω must be greater than the winding resistance, so something must be wrong with the generator measurement.

 

[jake jot]

What is meaning of o/c voltage? overclock? or open circuit voltage? or over charge? Also what is the s/c in "s/c current", shut current? step current?

I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms. I put the multimeter in AC instead of DC and tried hand cranking it so fast i don't mind if the rubber band would snap and highest I can get is 1.2 volts and 25mA. If it is set in DC, the voltage highest is 0.7 volts and 20mA. Why does putting multimeter in AC gives higher voltage?

Whereas the voltage (in generator mode) that can even make the rotor rotates by putting it sideways (to avoid friction) should be at least 7 volts dc and it's not rotating as fast. Why couldn't I get it to rotate at 1.2 volts? (i put it sideway to lessen friction from bottom).

What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?

I wonder if it has to do with the split commutator. Mine is the 2nd one.

13.7: Electric Generators and Back Emf - Physics LibreTexts

How do you turn a pulsed dc to direct dc? Maybe this was why I was only measuring 1.2 volts at most? (when in motor mode, the voltage that can get the rotate to start rotating must be at least 7 volts?)

 

[jake jot]

I wonder if it has to do with the split commutator. Mine is the 2nd one.

13.7: Electric Generators and Back Emf - Physics LibreTexts

View attachment 274632View attachment 274633

How do you turn a pulsed dc to direct dc? Maybe this was why I was only measuring 1.2 volts at most? (when in motor mode, the voltage that can get the rotate to start rotating must be at least 7 volts?)

Could the reason be the so called back emf? that is.. in motor, you need larger voltage to make it run and not proportional to the voltage in the output when it is in generator mode? I read in article above:

"
The generator output of a motor is the difference between the supply
voltage and the back emf. The back emf is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster, the
back emf grows, always opposing the driving emf, and reduces both the
voltage across the coil and the amount of current it draws. This effect is noticeable in many common situations. When a vacuum cleaner, refrigerator
, or washing machine is first turned on, lights in the same circuit dim briefly due to the IR drop produced in feeder lines by the large current drawn by the motor. "

 

[Tom.G]

The OP repeatd this query to me in a private msg. I chose to address it here publically.

Is it not if the rotor is a certain given speed. It should output the same voltage or current as when you use the battery of same voltage and current to turn the rotor (at same speed)?

When measuring as a motor and as a generator they should be measured at the same motor current. Under that condition they should be close to the same. If measured as a generator with little or no load they will be different. The lower voltage and current as generator indicate other, unknown, problems somewhere in the test setup.

Looking at the 1 A current with 9.8 V battery, the resistance (windings, connections, battery internal resistance) is less than 9.8 Ohm. How much less depends on the motor speed. Maybe your multimeter could measure the resistance at zero speed?

As far as I can see, 9.8 Ω must be greater than the winding resistance, so something must be wrong with the generator measurement.

WIth the stated wire size of 0.0078 inches, 9.8 Ohms indicates about 50 feet of wire, the 24 Ohms measured by the OP indicates about 130 to 150 feet on the winding. Bad commutator contact?

Overall, the various measurement inconsistencies point to both an instrumentation problem, such as an average-reading (or maybe peak-reading) meter being used for a pulsed measurement, and probably poor, bouncing, contact between the commutator and the brushes.

I suggest the OP find someone locally that can work on the actual device with him and may have more appropriate equipment and more hands-on experience. Trying to debug convoluted problems remotely with limited information is not very practical.

 

[jake jot]

The OP repeatd this query to me in a private msg. I chose to address it here publically.When measuring as a motor and as a generator they should be measured at the same motor current. Under that condition they should be close to the same. If measured as a generator with little or no load they will be different. The lower voltage and current as generator indicate other, unknown, problems somewhere in the test setup.WIth the stated wire size of 0.0078 inches, 9.8 Ohms indicates about 50 feet of wire, the 24 Ohms measured by the OP indicates about 130 to 150 feet on the winding. Bad commutator contact?

Overall, the various measurement inconsistencies point to both an instrumentation problem, such as an average-reading (or maybe peak-reading) meter being used for a pulsed measurement, and probably poor, bouncing, contact between the commutator and the brushes.

I suggest the OP find someone locally that can work on the actual device with him and may have more appropriate equipment and more hands-on experience. Trying to debug convoluted problems remotely with limited information is not very practical.

Please see message #97 and #98. Is it not caused by back EMF? Meaning you need to use larger voltage when used as generator to counter any back EMF?

 

[jake jot]

The OP repeatd this query to me in a private msg. I chose to address it here publically.When measuring as a motor and as a generator they should be measured at the same motor current. Under that condition they should be close to the same. If measured as a generator with little or no load they will be different. The lower voltage and current as generator indicate other, unknown, problems somewhere in the test setup.WIth the stated wire size of 0.0078 inches, 9.8 Ohms indicates about 50 feet of wire, the 24 Ohms measured by the OP indicates about 130 to 150 feet on the winding. Bad commutator contact?

Overall, the various measurement inconsistencies point to both an instrumentation problem, such as an average-reading (or maybe peak-reading) meter being used for a pulsed measurement, and probably poor, bouncing, contact between the commutator and the brushes.

I suggest the OP find someone locally that can work on the actual device with him and may have more appropriate equipment and more hands-on experience. Trying to debug convoluted problems remotely with limited information is not very practical.

Reading more about Back EMF. I think it's not the cause of the motor setup of the amazon generator requiring more voltage and current. Isn't it. Maybe it's the commutator contacts. I used a load that should light up if 3v or more and it still could never light up.

It's not good idea to bring the amazon generator for kid into a university and ask the professor when I wasn't enrolled in the school. And technician at the washing machine motor rewinding shop may not want to look at my little amazon motor/generator or the power company.

Anyway I just want to understand the basic concepts and I think it's all enough for me.

One last question I need to know to conclude the thread is the following.

In pole attraction motor like brushless motors or the amazon motor where the solenoid with iron core creates north-south poles. So the purpose of commutator is to make sure it is north or south at one side so the rotor continues to rotate. I understand this concept well. And the purpose of the current is to produce the poles electromagnetically and make it switch or maintain north, south via commutator. I need to know the following thought experiment to make sure I understand the basic concept.

Supposed you had a pole attraction motor where the rotor has one end a monopole that switches north and south, and the other end also a monopole that switches north and south opposite to the other. Will the motor runs continuously without any current? So it will keep rotating because of Newton's 1`st law which states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it?

Kindly answer the above and not say it is bad example. Because this whole thread is to understand the above thought experiment. Thank you.

 

[Merlin3189]

Sorry, just saving typing.
o/c = open circuit. ie. negligible current, only voltage. Voltmeter has very high resistance, 20 kΩ - 10 MΩ
s/c = short circuit. ie negligible voltage, only current. Ammeter has very low resistance.

"I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms."
So this should be correct and your other measurements must have faults.

"Why does putting multimeter in AC gives higher voltage?"
Measuring DC is easy: there is only one value.
AC is more confusing: you could measure the peak value, the RMS value (which is equivalent to DC value in resistive circuit), or average value (average modulus, otherwise it would be zero for pure AC!) It gets even more difficult when the voltage is a mixture of AC and DC. What you see depends on your meter. Many meters give an average, but modern electronic meters can do RMS (and maybe peak as well.)
But, for all that, I can't see how you get the values you do.

"What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?"

This is a model circuit for a DC motor. There is a source of emf, whose value depends on the speed of rotation and the motor constant - a function of the stator magnetic field and rotor windings.
At a given speed the emf is determined, kω. The current is then determined by the internal resistance R and the external load resistance. The voltage you measure at the output terminals or across the load will be less than Bemf by the voltage across the internalresistance R. If there is no load and no current, then you measure the Bemf.

When you drive it as a motor, the current is determined by the resistance, the applied voltage and the speed that it rotates - which will depend on the mechanical load and friction.

Say, at 1000 rpm the emf is 5 V (and R is 24 Ω as you measured) then attaching a 26 Ω load will give a current of $\frac {5 V}{24 +26 Ω } = 0.1 A$ with $0.1 A \times 26 Ω = 2.6 V$ across the load.

To drive it as a motor you will need to apply more than 5 V to get that same 1000 rpm, because you need some current to give the torque to overcome friction.
If you apply 5 V, then $\frac {5V}{24 Ω } = 208 mA$ will flow (neglecting battery resistance.)
If the torque that generates is enogh to overcome friction, then the rotor will accelerate.
That will cause some Bemf which will reduce the current. For eg. if the rotor got up to 900 rpm, the Bemf will be 4.5 V, so the current will drop to $\frac {5-4.5 V}{24 Ω } = 21 mA$
If that is just enough to overcome friction at that speed, then that's the speed it will run at. If it isn't, then it won't ever reach that speed.

If for eg. it needs about 25 mA to have the torque needed to overcome friction, that requires a voltage across R of $0.025 A \times 24 Ω = 0.6 V$
So with 5 V battery the Bemf can rise to 4.4 V , leaving 0.6 V across R. But Bemf = 4.4 V means it is rotating at $1000 \times \frac {4.4}{5} = 880 rpm$

If you now add a mechanical load, that requires extra torque, which requires extra current.
Say eg. the extra load requires a current of 100 mA. Now the total required current is 125 mA to give the torque needed for the load and friction.
That requires a voltage across R of $0.125 A \times 24 Ω = 3 V$
So now we can afford only 2 V of Bemf and the motor must slow to $1000 \times \frac{2}{5) = 400 rpm$

If we wanted the motor to run at 1000 rpm with that load, then we need our 125 mA current to provide the correct torque and enough voltage to overcome the 5 V Bemf. So with 125 mA needing 3 V across R, we need a supply of 8 V.

That's probably laboured the point a bit, but I hope it shows the nature of the model and the source of the values.

 

[jake jot]

Sorry, just saving typing.
o/c = open circuit. ie. negligible current, only voltage. Voltmeter has very high resistance, 20 kΩ - 10 MΩ
s/c = short circuit. ie negligible voltage, only current. Ammeter has very low resistance.

"I measured the resistance of the winding of the amazon generator by directly tapping the leads at the 2 commutators. It is exactly 24 ohms."
So this should be correct and your other measurements must have faults.

"Why does putting multimeter in AC gives higher voltage?"
Measuring DC is easy: there is only one value.
AC is more confusing: you could measure the peak value, the RMS value (which is equivalent to DC value in resistive circuit), or average value (average modulus, otherwise it would be zero for pure AC!) It gets even more difficult when the voltage is a mixture of AC and DC. What you see depends on your meter. Many meters give an average, but modern electronic meters can do RMS (and maybe peak as well.)
But, for all that, I can't see how you get the values you do.

"What I exactly want to know is this principle. Whether if the rotor at motor mode rotates at a certain RPM at a given voltage and current. Whether if the rotor is put in generator mode. It will output the same voltage and current at the same RPM. What do you think?"

View attachment 274676This is a model circuit for a DC motor. There is a source of emf, whose value depends on the speed of rotation and the motor constant - a function of the stator magnetic field and rotor windings.
At a given speed the emf is determined, kω. The current is then determined by the internal resistance R and the external load resistance. The voltage you measure at the output terminals or across the load will be less than Bemf by the voltage across the internalresistance R. If there is no load and no current, then you measure the Bemf.

When you drive it as a motor, the current is determined by the resistance, the applied voltage and the speed that it rotates - which will depend on the mechanical load and friction.

Say, at 1000 rpm the emf is 5 V (and R is 24 Ω as you measured) then attaching a 26 Ω load will give a current of $\frac {5 V}{24 +26 Ω } = 0.1 A$ with $0.1 A \times 26 Ω = 2.6 V$ across the load.

To drive it as a motor you will need to apply more than 5 V to get that same 1000 rpm, because you need some current to give the torque to overcome friction.
If you apply 5 V, then $\frac {5V}{24 Ω } = 208 mA$ will flow (neglecting battery resistance.)
If the torque that generates is enogh to overcome friction, then the rotor will accelerate.
That will cause some Bemf which will reduce the current. For eg. if the rotor got up to 900 rpm, the Bemf will be 4.5 V, so the current will drop to $\frac {5-4.5 V}{24 Ω } = 21 mA$
If that is just enough to overcome friction at that speed, then that's the speed it will run at. If it isn't, then it won't ever reach that speed.

If for eg. it needs about 25 mA to have the torque needed to overcome friction, that requires a voltage across R of $0.025 A \times 24 Ω = 0.6 V$
So with 5 V battery the Bemf can rise to 4.4 V , leaving 0.6 V across R. But Bemf = 4.4 V means it is rotating at $1000 \times \frac {4.4}{5} = 880 rpm$

If you now add a mechanical load, that requires extra torque, which requires extra current.
Say eg. the extra load requires a current of 100 mA. Now the total required current is 125 mA to give the torque needed for the load and friction.
That requires a voltage across R of $0.125 A \times 24 Ω = 3 V$
So now we can afford only 2 V of Bemf and the motor must slow to $1000 \times \frac{2}{5) = 400 rpm$

If we wanted the motor to run at 1000 rpm with that load, then we need our 125 mA current to provide the correct torque and enough voltage to overcome the 5 V Bemf. So with 125 mA needing 3 V across R, we need a supply of 8 V.

That's probably laboured the point a bit, but I hope it shows the nature of the model and the source of the values.

Thanks for the details.

One full hand crank revolution at bottom is equal to 5 rotor full turn. How do you compute the RPM based on a normal hand effort to turn it? (anyway I measured 2 revolution per sec tops when I turned the hand crank). So in one minute there are 2 x 60 x 5 = 600 RPM or make it 500 RPM since I can't maintain full 2 revolution per sec in one minute continuously.)

The winding resistance is 24 ohms, and it measured 1 A when connected to 9.8v battery in motor mode. What should be the output in generator mode based on these data? Assuming there is no load in either the generator and motor mode (except the meter and batteries).

 

[Merlin3189]

V=IR so 9.8 V = 1 A x 24 Ω means something is not right.
It should need 24 V to drive 1 A through 24 Ω resistance, even when it's not moving. When it moves, it should require even more voltage.

As someone else said, it's hard to guess which bit is wrong when you're not there to see exactly what's going on. My guess would be to distrust the resistance reading. Have you checked the meter shows zero when you short the test leads together?
Otherwise, the 1 A has to be too high.

 

[jake jot]

V=IR so 9.8 V = 1 A x 24 Ω means something is not right.
It should need 24 V to drive 1 A through 24 Ω resistance, even when it's not moving. When it moves, it should require even more voltage.

As someone else said, it's hard to guess which bit is wrong when you're not there to see exactly what's going on. My guess would be to distrust the resistance reading. Have you checked the meter shows zero when you short the test leads together?
Otherwise, the 1 A has to be too high.

I'm recharging the 8 rechargeable batteries.

But in other loads, 8 of these can make 1A, isn't it?

I'll try to use other multi tester. I need to get to the bottom of it. Because after determining why it gives such. I'd put the amazon motor and generator in my cabinet and I'd not touch them again for my whole life. So this is one time to figure it out.

Also the product at amazon is now available. It's the only hand crank available for about 7 billion humans. So maybe you can try to purchase one and measure it using your multimeter? For science. For education. For humanity.

Amazon.com: EUDAX School DIY Dynamo Lantern Educational STEM Building ,Labs Demonstration Motor Activity Teaching Model Hand Cranked Power Electricity DC Electric Generator Physical Science Experiment Education: Toys & Games

 

[jake jot]

V=IR so 9.8 V = 1 A x 24 Ω means something is not right.
It should need 24 V to drive 1 A through 24 Ω resistance, even when it's not moving. When it moves, it should require even more voltage.

As someone else said, it's hard to guess which bit is wrong when you're not there to see exactly what's going on. My guess would be to distrust the resistance reading. Have you checked the meter shows zero when you short the test leads together?
Otherwise, the 1 A has to be too high.

I borrowed a meter but it's an ac clump meter with current in AC. But it has ohm meter and I measured the same 24 ohms. Using my multimeter. I tried to measure the current again. I noticed it is always negative (-) even if i changed the terminals. Also the amperage was not steady. So I guess it just can't read the ampere of it properly because the waveform is not really pure dc but something like the following? But I wonder why it always displays negative no matter how i changed the leads. When I changed the current to AC. It displays even higher reading (double).

What is the shape of the current for the above? Is it sinusoidal, like the above, or just plain flat line dc?

 

[Merlin3189]

It should be something like your graph, a full wave rectified sinewave. I'd reckon a bit peakier, with the sine shape being distorted by non-uniform field and the nodes being distorted by a dead spot as the brushes are either in a gap or are shorting both sectors.

" It's the only hand crank available for about 7 billion humans. "
I'm sure I could build one myself. Maybe a bit more refined than the bits of wire I put together for your first experiment - though that did work. I think Meccano might be suitable.

I'm not quite sure what I'd be trying to find out though? These sort of motors are pretty well understood, so you can work out what should happen a lot easier than you can measure it. And lazy people like me can get all the info we want off the internet.

If you really want to measure things, this is not an easy project. A simple bit of equipment like this demo device is probably intended to show qualitative effects. Friction, resistance and efficiency are not the key factors in its design.

 

[jake jot]

It should be something like your graph, a full wave rectified sinewave. I'd reckon a bit peakier, with the sine shape being distorted by non-uniform field and the nodes being distorted by a dead spot as the brushes are either in a gap or are shorting both sectors.

" It's the only hand crank available for about 7 billion humans. "
I'm sure I could build one myself. Maybe a bit more refined than the bits of wire I put together for your first experiment - though that did work. I think Meccano might be suitable.

I'm not quite sure what I'd be trying to find out though? These sort of motors are pretty well understood, so you can work out what should happen a lot easier than you can measure it. And lazy people like me can get all the info we want off the internet.

If you really want to measure things, this is not an easy project. A simple bit of equipment like this demo device is probably intended to show qualitative effects. Friction, resistance and efficiency are not the key factors in its design.

Ok. I just want to understand the basic concepts and I think it's all enough for me.

One last question I need to know to conclude the thread is the following.

In pole attraction motor like brushless motors or the amazon motor where the solenoid with iron core creates north-south poles. So the purpose of commutator is to make sure it is north or south at one side so the rotor continues to rotate. I understand this concept well. And the purpose of the current is to produce the poles electromagnetically and make it switch or maintain north, south via commutator. I need to know the following thought experiment to make sure I understand the basic concept.

Supposed you had a pole attraction motor where the rotor has one end a monopole that switches north and south, and the other end also a monopole that switches north and south opposite to the other. Will the motor runs continuously without any current? So it will keep rotating because of Newton's 1`st law which states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it?

Kindly answer the above and not say it is bad example. Because this whole thread is to understand the above thought experiment. Thank you.

 

[Merlin3189]

Supposed you had a pole attraction motor where the rotor has one end a monopole that switches north and south, and the other end also a monopole that switches north and south opposite to the other. Will the motor runs continuously without any current? So it will keep rotating because of Newton's 1`st law which states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it?

? Well, AFAIK monopoles don't exist, but if you had monopoles at each end which were always opposite polarity, that would be a dipole, which does exist !
Putting such niceties aside, you seem to be describing our little commutated DC motor, except that you want the rotor magnet to reverse without any input. I don't think that will happen. Either you need current to make it a magnet at all, of whichever polarity, or the remnance which keeps ferromagnetic materials magnetised after you remove the current, leaves it permanently magnetised in that polarity. Then you need a current to reverse the polarity. So no free lunch as far as I can see.
Newton's third law is irrelevant because there are forces acting on the rotor. If it has magnetic poles, they are attracted or repelled by the stator poles and that's forces. If you get rid of all the magnets, then you just end up with a flywheel. Well designed with low friction, that could continue spinning for some time, but would gradually slow down. It's a basic law of nature for macroscopic objects. If you want to see perpetual motion, maybe you need to look at the subatomic level (so don't ask me!) Some particles seem to have permanent spin, though whether these Physicist types mean that in the same sense as I do when talking about motors, I doubt.

 

[jake jot]

? Well, AFAIK monopoles don't exist, but if you had monopoles at each end which were always opposite polarity, that would be a dipole, which does exist !
Putting such niceties aside, you seem to be describing our little commutated DC motor, except that you want the rotor magnet to reverse without any input. I don't think that will happen. Either you need current to make it a magnet at all, of whichever polarity, or the remnance which keeps ferromagnetic materials magnetised after you remove the current, leaves it permanently magnetised in that polarity. Then you need a current to reverse the polarity. So no free lunch as far as I can see.
Newton's third law is irrelevant because there are forces acting on the rotor. If it has magnetic poles, they are attracted or repelled by the stator poles and that's forces. If you get rid of all the magnets, then you just end up with a flywheel. Well designed with low friction, that could continue spinning for some time, but would gradually slow down. It's a basic law of nature for macroscopic objects. If you want to see perpetual motion, maybe you need to look at the subatomic level (so don't ask me!) Some particles seem to have permanent spin, though whether these Physicist types mean that in the same sense as I do when talking about motors, I doubt.

Before I put these in box and stored in the attic. I need to make a little project. I want to make the solenoid so thin it will be purely pole attraction. Remember earlier we were not sure whether it is:

"Force Exerted on a Current-Carrying Conductor in a Magnetic Field" or

"Force Exerted by Like or Opposing Poles of Magnets"

maxwell_demons gave us the distinctions. Although the amazon generator is of the former mechanism.

Do you know commercially available solenoids so thin I can just add it above instead of building one?

 

[jake jot]

? Well, AFAIK monopoles don't exist, but if you had monopoles at each end which were always opposite polarity, that would be a dipole, which does exist !
Putting such niceties aside, you seem to be describing our little commutated DC motor, except that you want the rotor magnet to reverse without any input. I don't think that will happen. Either you need current to make it a magnet at all, of whichever polarity, or the remnance which keeps ferromagnetic materials magnetised after you remove the current, leaves it permanently magnetised in that polarity. Then you need a current to reverse the polarity. So no free lunch as far as I can see.
Newton's third law is irrelevant because there are forces acting on the rotor. If it has magnetic poles, they are attracted or repelled by the stator poles and that's forces. If you get rid of all the magnets, then you just end up with a flywheel. Well designed with low friction, that could continue spinning for some time, but would gradually slow down. It's a basic law of nature for macroscopic objects. If you want to see perpetual motion, maybe you need to look at the subatomic level (so don't ask me!) Some particles seem to have permanent spin, though whether these Physicist types mean that in the same sense as I do when talking about motors, I doubt.

I was looking for a thin solenoid in the net to power the motor (see last message) to prove pole attraction is the mechanism and found most have iron cores. It seems this is required to increase the magnetic flux. I wonder if my motor would even run if there is no iron core. I read about about solenoid and rules of windings. What size of wire is the most useful in your experience so I can maybe buy a small roll of it.

But what perplexed me so much was when I came across the so called Permanent Magnet DC motor. I watched most youtube videos about it and read many articles. But there is none that explains it clearly as the ordinary iron core dc motors.

What are coreless DC motors? (motioncontroltips.com)

Do you know a site that explains it. After 3 hours of googling and reading dozens of articles and videos. There is none that explains it clearly and I can't seem to fathom it. How does it work compared to the typical iron core motors?

 

[jake jot]

I was looking for a thin solenoid in the net to power the motor (see last message) to prove pole attraction is the mechanism and found most have iron cores. It seems this is required to increase the magnetic flux. I wonder if my motor would even run if there is no iron core. I read about about solenoid and rules of windings. What size of wire is the most useful in your experience so I can maybe buy a small roll of it.

But what perplexed me so much was when I came across the so called Permanent Magnet DC motor. I watched most youtube videos about it and read many articles. But there is none that explains it clearly as the ordinary iron core dc motors.

What are coreless DC motors? (motioncontroltips.com)

View attachment 274888

Do you know a site that explains it. After 3 hours of googling and reading dozens of articles and videos. There is none that explains it clearly and I can't seem to fathom it. How does it work compared to the typical iron core motors?

My dilemma started when I planned to get Neodymium core, but I read these are permanent magnets. We know that in the motor iron core beneath the winding, the direction of flux changes at every commutator contact changes. So how can a permanent magnet worked there I wondered. Then I read this site.

Permanent magnet Vs. iron cored DC motors (eurekamagazine.co.uk)

This paragraph confused me:

"Like the iron core magnet the conductors themselves produce a magnetic field which is not always contributing to the field of the permanent magnet and superimpose themselves on the existing magnetic field. This weakens the field of the permanent magnet and is avoided by increasing the air gap between the magnets and the coil. Increasing the air gap however means the flux density is reduced which in turn effects the amount of torque the motor can produce."

Then I read and watched videos about Permanent magnet dc motors. Now I think what confused me was the article above was mixing the concepts. It mentioned Permanent magnet but it is not the same as the Permanent Magnet DC motor. This line can clarify it.

What is a conventional DC motor? Isn't our amazon motor a conventional DC motor? I thought conventional DC motor used magnets. So they don't? And therefore my amazon motor belonged to the category of Permanent Magnet DC Motor?

If so, then what I wanted to understand was the so called coreless DC motor after all, which the first article above mentioned was the same as permanent dc motor. Hence the confusions.

In this site How to Select a DC Motor: Coreless and Iron Core Brushed DC Motors | PTE (powertransmission.com)

The permanent magnet within the rotor doesn't move along with it, correct? Because unless it is brushless dc motor, I can't fathom how you can replace the iron core with permanent north and south magnet and still work with commutator (note the above still has commutator, you don't need commutator and brushes in brushless dc motor but it needs complex circuity to synchronize the current in the stator).

So in the following description quote earlier and quoting again for emphasis, the permanent magnet inside the rotor doesn't move with the rotor, right? If it still moves, then I still don't get it how it works. What is this describing then?

"Like the iron core magnet the conductors themselves produce a magnetic field which is not always contributing to the field of the permanent magnet and superimpose themselves on the existing magnetic field. This weakens the field of the permanent magnet and is avoided by increasing the air gap between the magnets and the coil. Increasing the air gap however means the flux density is reduced which in turn effects the amount of torque the motor can produce."

Merlin3189 and other experts in motors, what you think? Thank you.

 

[Merlin3189]

This is an interesting sort of motor. I only came across it quite recently, when looking for a low inertia motor.
It is very definitely a "wire in magnetic field" because there is no rotating iron/magnet.
The field is generated by the fixed core magnet, which is inside the rotor, but doesn't rotate with it. (You could have the magnet outside, but then it would be bigger and Neodymium is expensive.)
The rotor is made up of lots of coils, as you can see from the multi-sector commutator. The exact way of winding these, I'm still looking for, but it looks as if each coil is twisted and overlaps the adjacent coils. This presumably helps make the construction rigid, but I think also smooths the torque.
Because there is just the coils with no iron, the mass and moment of inertia of the rotor is much lower than in a traditional motor. So the motor accelerates and decelerates much quicker.

They also mention here (something I'd not thought of) that it is more efficient, because you are not continually reversing the field inside iron (or other ferromagnet.) The magnet and iron/ferrite shell are all static and have a more or less static field.
In an iron cored rotor the polarity of the magnetisation is reversed every half cycle by the commutation. When you do this, there is hysteresis and energy is lost on each cycle. You can read up on this, if you don't know about it.

The magnet and shell are joined at the far end. The coil pushes into the gap from the near end. (sorry I've missed out the axle hole in the centre of the magnet.) The coil is like a cylinder open at the far end, with the commutator at the near end. The coil is encapsulated in resin for rigidity.
The windings run on a helical path for half a turn along the sides of the coil, then double back for the second half. The next turn is slightly offset (by the width of the wire). The diagram says it is a Felix Faulhaber design, so you can look it up, if you've a better brain for 3D contortions than mine. I gather it gives a more even torque than (relatively) simple paraxial coils.

 

[jake jot]

This is an interesting sort of motor. I only came across it quite recently, when looking for a low inertia motor.
It is very definitely a "wire in magnetic field" because there is no rotating iron/magnet.
The field is generated by the fixed core magnet, which is inside the rotor, but doesn't rotate with it. (You could have the magnet outside, but then it would be bigger and Neodymium is expensive.)
The rotor is made up of lots of coils, as you can see from the multi-sector commutator. The exact way of winding these, I'm still looking for, but it looks as if each coil is twisted and overlaps the adjacent coils. This presumably helps make the construction rigid, but I think also smooths the torque.
Because there is just the coils with no iron, the mass and moment of inertia of the rotor is much lower than in a traditional motor. So the motor accelerates and decelerates much quicker.

They also mention here (something I'd not thought of) that it is more efficient, because you are not continually reversing the field inside iron (or other ferromagnet.) The magnet and iron/ferrite shell are all static and have a more or less static field.
In an iron cored rotor the polarity of the magnetisation is reversed every half cycle by the commutation. When you do this, there is hysteresis and energy is lost on each cycle. You can read up on this, if you don't know about it.

View attachment 274927The magnet and shell are joined at the far end. The coil pushes into the gap from the near end. (sorry I've missed out the axle hole in the centre of the magnet.) The coil is like a cylinder open at the far end, with the commutator at the near end. The coil is encapsulated in resin for rigidity.
The windings run on a helical path for half a turn along the sides of the coil, then double back for the second half. The next turn is slightly offset (by the width of the wire). The diagram says it is a Felix Faulhaber design, so you can look it up, if you've a better brain for 3D contortions than mine. I gather it gives a more even torque than (relatively) simple paraxial coils.

About "In an iron cored rotor the polarity of the magnetisation is reversed every half cycle by the commutation". I'm familiar with the whole concept and the purpose of the commutator and recently the iron core and magnetization. I have spent a whole week working with the commutators in my amazon motor and generators. That's the whole purpose of them, to have hands on experience.

In the coreless dc motor. If the magnet inside it is fixed, then it all makes sense. I just can't fathom a permanent magnet in rotor that moves unless it's a brushless dc motor with current synchronization circuitry.

So the permanent magnet inside the rotor acts just like stator with the winding rotating outside with respect to it. This is the punchline of it right?

What is the size of the winding that is the most useful? so I can buy it now and try different winding length in my amazon motor and generator to see how they would behave.

The last time I tried to wind coils was 30 years ago.

At that time, i was finding out the claim about the so called scalar wave, which is allegedly produced when one cancels magnetic field. So I wired the above in photo according to the instructions.

What I found out was simply when one separated the 2 wires (they power a lamp with AC source). There were less cancellation of the magnetic field and this disturbed the body sentient bioetheric field. It's the AC 60 Hertz frequency not friendly. When the wire was put together, the magnetic field was less so less disturbance. This is all there is to it. So much for scalar waves.

 

[Tom.G]

This is an interesting sort of motor. I only came across it quite recently, when looking for a low inertia motor.
It is very definitely a "wire in magnetic field" because there is no rotating iron/magnet.

Another advantage of those "Ironless Rotor" motors is they don't cog when run at slow speeds.

They also do not have a "detent action" when powered down and manually rotated. This was crucial in one project that had a "lighthouse" style rotating LASER beam. There were occassions where the motor would have to be stopped and the beam aimed manually at a specific spot. That was not possible with a motor core that had Iron in it.

Cheers,
Tom

 

[jake jot]

Another advantage of those "Ironless Rotor" motors is they don't cog when run at slow speeds.

They also do not have a "detent action" when powered down and manually rotated. This was crucial in one project that had a "lighthouse" style rotating LASER beam. There were occassions where the motor would have to be stopped and the beam aimed manually at a specific spot. That was not possible with a motor core that had Iron in it.

Cheers,
Tom

Without iron core, the magnetic field induced in the rotor is weaker. I will try to wind my motor without iron core in the rotor. I don't know if it will run or need much higher current.

Now in the coreless dc motor. Wont the rotor induced magnetic field be poorer without any core. Then how does it differ to conventional dc motor without any iron cores as far as the rotor in concerned?

 

[Tom.G]

Wont the rotor induced magnetic field be poorer without any core.

Yes it will. You will need more turns and/or more current for the same torque due to the weaker magnetic field.

Cheers,
Tom

 

[jake jot]

View attachment 274847

Before I put these in box and stored in the attic. I need to make a little project. I want to make the solenoid so thin it will be purely pole attraction. Remember earlier we were not sure whether it is:

"Force Exerted on a Current-Carrying Conductor in a Magnetic Field" or

"Force Exerted by Like or Opposing Poles of Magnets"

maxwell_demons gave us the distinctions. Although the amazon generator is of the former mechanism.

Do you know commercially available solenoids so thin I can just add it above instead of building one?

I finally did my second amazon motor modification to test pole attraction by making the rotor winding thinner than the original.

I used nail and steel screw as core (because I couldn't find 2 nails). Need these to wrap the AWG #28 windings (4 times) around the shalf center. This is video of it running.


I initially used 2 AA batteries, it didn't run unlike the original amazon rotor. I then tried to use 4 batteries, It still won't run. I need 6 batteries to make it run.

Does the above prove the principle of "Force Exerted by Like or Opposing Poles of Magnets". Or do you still think there is possibility of "Force Exerted on a Current-Carrying Conductor in a Magnetic Field"?

Say, when doing winding, what happens if say one side of rotor has more winding (say double)? because I couldn't make them totally equal in length. It's tiring to rewind so I'd not make any more project, hence just need to know what would theoretically happens, like how it would behave if the north is more powerful than the south (is this possible)?.

Thanks guys!

 

[jake jot]

Yes. The iron shell guides the flux from the magnet at the top and produces a horizontal field around the rotor.
Your first diagrams used two magnets one on each side of the rotor. The field can be strengthened by joining the outside faces with iron, completing the manetic "circuit".
Early motors sometimes used horseshoe magnets to have the same effect.
This idea is very economical because you need only one simple magnet, then the cheap shaped piece of iron brings the field to where you want it.

Yes. The iron shell guides the flux from the magnet at the top and produces a horizontal field around the rotor.
Your first diagrams used two magnets one on each side of the rotor. The field can be strengthened by joining the outside faces with iron, completing the manetic "circuit".
Early motors sometimes used horseshoe magnets to have the same effect.
This idea is very economical because you need only one simple magnet, then the cheap shaped piece of iron brings the field to where you want it.

I was trying to commute at different locations in the last project above (see last message). It can rotate when commuted at horizontal. When trying to commute vertically. It won't and I noticed the configuration of the amazon motor was not the typical ones shown at youtube. You mentioned above about "The iron shell guides the flux from the magnet at the top and produces a horizontal field around the rotor."

But how come when I used a magnet of same polarity that repelled the top magnet, it didn't do likewise in the lower metal connected to it? Then I tried this experiment:



Without the metal, I held the the magnets at the top and it rotated.

Do you have illustrations how the iron shell guides the flux from the magnet? Why doesn't it have any polarity as putting magnetic north at the metal won't repel it?

Now pertaining to your illustration at msg #45

I know commuting at vertical won't rotate the rotor. But when you have north and south pole magnets put at top without any metal to extend the fluxes (this won't create any north or south in the metal itself). The analysis is different than your illusration but still understandable. One can just imagine you lower bottom left side as the top.

What motor used the same magnets at top like the amazon motor? It proves the principle of "Force Exerted by Like or Opposing Poles of Magnets". Or do you still think there is possibility of "Force Exerted on a Current-Carrying Conductor in a Magnetic Field"?