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Increasing motor's RPM + Torque with stronger magnetic fields?

Dretron

Jun 9, 2012
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Hallo,

I noticed something a few weeks back when I was playing around with magnets and a motor.

When I added a stronger magnet to a certain position in the motor the rpm would suddenly increase! At the same input of power! Added another magnet and it becomes lounder!

A demonstration here!

Now I've researched countless times and found magnetic force on a stator/loop = IL x B

If the magnetic field is strengthen there is more force on the stator and more torque is generated and speed will increase with the same input.

Or even changing the magnet type from ceramic to neodymium(strongest kind) would make a significant change!


Interesting phenomena! Now the thing is... I'm skeptical about is this whole process... What I've learned in physics made me more negative to theses things... And I feel there is something missing/wrong/leading to failure of some sort... One problem was balancing the magnetic field: Easily solvable.
Other problem was friction within the bearings: Can be solved as well...
Also CEMF is considerable but its not really a problem isn't it?

So... What other problems can this due for me? I don't want to get exited and study this and eventually its pointless...

Rpm is not proportional to torque I know that... If one increases the other decreases... But in this case it seems to me like both are evenly increasing... Since were causing a temporary dipole(stator which has current flowing = temporary) to be repelled and attracted by the permanent dipoles' magnetic field... It makes some sense but again I'm skeptical of this effect.

Thanks everyone.

Dre,
 

BobK

Jan 5, 2010
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Did you measure the current being drawn? Voltage is not the same as power. Voltage times current = power.

Bob
 

john monks

Mar 9, 2012
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You are probably reducing the strength of the field magnet causing the motor to have to spead up to make up for the lost counter emf of the motor.
If you want I can go into this in detail.
 

Dretron

Jun 9, 2012
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Did you measure the current being drawn? Voltage is not the same as power. Voltage times current = power.

Bob

I understand Voltage is not as same as power. I see voltages as the pressure of water and current is that water (simple example).

Didn't test the current in with my set up! I will do it and come back.

One thing though if I tested the current and it was the same thing... Is that explainable?
 
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Dretron

Jun 9, 2012
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You are probably reducing the strength of the field magnet causing the motor to have to spead up to make up for the lost counter emf of the motor.
If you want I can go into this in detail.


Good point! Since increasing a magnetic field would generally make a higher torque so by increasing speed you're making a weaker magnetic field... Makes sense but please go on!

I've been looking for useful info. And experts to discuss with.

Dre,
 

Dretron

Jun 9, 2012
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You are probably reducing the strength of the field magnet causing the motor to have to spead up to make up for the lost counter emf of the motor.
If you want I can go into this in detail.

Why is it when the magnetic strength is greater... The torque is higher by the speed is not... Its confusing...

If one would increase magnetic strength that means the rpm would be less... And torque would be greater? But the rpm isn't reliant on the magnetic field/force/strength etc..

Please give detail's to you're answers so I could understand.
Thank you,

Dre.
 
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john monks

Mar 9, 2012
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Let me use an analogy.
Let's say that we had a very strong magnetic field of one telsa pointing straight up through a table one meter square. Now if you had a wire one meter long, stretched it with your hands and transferased it at an even rate across the table in a direction perpendicular to the length of the wire you will develop one volt of electricity from end to end across the length of the wire.
Now suppose you connect the same wire with a one volt source the wire will move across the table at a rate of one meter per second.
Now let's double the voltage and the wire will move at a rate of two meters per second.
And just moving the wire with your hands at twice the rate you will get double the voltage or two volts from end to end.
Now let's try reducing the magnetic force to one half tesla. Now moving the wire at one meter per second you will only get one half volts.
Now if you place one half volts on the wire it will move at a rate of one meter per second.
But now let's try placing two volts on the same wire with only one half tesla coming out of the table the wire will move at a rate of two meters per second.
Now if we go back to the table with one tesla and the one meter long wire. If we connect a one ohm resistor across the ends of the wire and is placed some distance from the magnetic field you will have to push with a force of one newton to move the wire one meter per second as before. There will be one volt generated as before and there will be one ampere of current traveling through the wire and the resistor.
Now let's place a one volt battery in series with the resistor and let go of the wire. It will travel at a rate of one meter per second. If you hold the wire still you will be applying a force of one newton.
So applying a load or a one ohm resistor to the moving wire causes a force to be applied to the wire.
And with no load there will be no force on the wire. This is because of counter EMF.

A similar thing is taking place in the motor.
This effect is because of what is called counter emf or the electrical voltage generated by wire passing through a magnetic field.
So in a sense your motor is becoming a generator. If you only have half the magnetic field coming from tha magnets you will have to spin the motor twice as fast for the same voltage.

A perfect motor will draw no current once the motor reaches it's maximum speed. Motors draw current because of internal mechanical resistance. And when the motor runs faster more power is required to turn the rotor. So I would expect the current to increase when you apply a magnet as you did before because of the extra power required. Measure the current and find out.

If you understand this concept thoroughly I'm sure you will become a great electrical engineer.
There's no way you could have asked a better question. You did an experiment, got a result, and now you can start formulating theories.
 
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(*steve*)

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I'm going to have to read through this one again myself...
 

Dretron

Jun 9, 2012
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Let me use an analogy.
Let's say that we had a very strong magnetic field of one telsa pointing straight up through a table one meter square. Now if you had a wire one meter long, stretched it with your hands and transferased it at an even rate across the table in a direction perpendicular to the length of the wire you will develop one volt of electricity from end to end across the length of the wire.
Now suppose you connect the same wire with a one volt source the wire will move across the table at a rate of one meter per second.
Now let's double the voltage and the wire will move at a rate of two meters per second.
And just moving the wire with your hands at twice the rate you will get double the voltage or two volts from end to end.
Now let's try reducing the magnetic force to one half tesla. Now moving the wire at one meter per second you will only get one half volts.
Now if you place one half volts on the wire it will move at a rate of one meter per second.
But now let's try placing two volts on the same wire with only one half tesla coming out of the table the wire will move at a rate of two meters per second.
Now if we go back to the table with one tesla and the one meter long wire. If we connect a one ohm resistor across the ends of the wire and is placed some distance from the magnetic field you will have to push with a force of one newton to move the wire one meter per second as before. There will be one volt generated as before and there will be one ampere of current traveling through the wire and the resistor.
Now let's place a one volt battery in series with the resistor and let go of the wire. It will travel at a rate of one meter per second. If you hold the wire still you will be applying a force of one newton.
So applying a load or a one ohm resistor to the moving wire causes a force to be applied to the wire.
And with no load there will be no force on the wire. This is because of counter EMF.

A similar thing is taking place in the motor.
This effect is because of what is called counter emf or the electrical voltage generated by wire passing through a magnetic field.
So in a sense your motor is becoming a generator. If you only have half the magnetic field coming from tha magnets you will have to spin the motor twice as fast for the same voltage.

A perfect motor will draw no current once the motor reaches it's maximum speed. Motors draw current because of internal mechanical resistance. And when the motor runs faster more power is required to turn the rotor. So I would expect the current to increase when you apply a magnet as you did before because of the extra power required. Measure the current and find out.

If you understand this concept thoroughly I'm sure you will become a great electrical engineer.
There's no way you could have asked a better question. You did an experiment, got a result, and now you can start formulating theories.

Ow ok, so due to C-EMF this would cause the motor to require more current. So the magnetic field is some what creating a major resistance as it's field strengthens.


However, a magnetic field's are the cause of rotational motion...
If the C-EMF, should make my input less?

I'll test things again and be back!
 

Dretron

Jun 9, 2012
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Another thing to add:

When the motor is speeding up that the rpm is significantly high, due to the C-EMF there will be less current drawn in.Wouldn't that contradict the idea of higher input current? That would mean the speed will decrease... Torque would increase.

Because when a motor is loaded its speed will eventually decrease and more input is needed to continue doing work on the load. When more magnets are applied more torque and speed are added up.

(Thats all my theory, I would test this as soon as I get my part back.)

Ow + its a known fact in efficiency to have stronger magnets like neodymium so that could help my "theory" a bit.

Since were looking at both RPM & Torque. I should measure both the Voltage input and Current input before and after the addition of stronger magnets.

Dre,
 
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Dretron

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Again... When I think about this matter even more...
More ideas are coming up about torque and rpm the relationship of the two.


Let me use an example to illustrate my thought:

I have an electric motor and a heavy wheel acting as the load.

Now when the electric motor is turned on without the wheel attached as a load, it will instantly run easily and go fast because there is no mechanical resistance of the load. So the motor would eventually speed up and not need more current because of the C-EMF. We can agree that the hardest part was only the beginning.

Now when I attach a load. That is the wheel... The motor starts to draw more and more current for torque! And the RPM significantly drops and the C-EMF would be less generated.However, if the "LOAD" was not a major "resistance" to the motor... In a sense the motor can "handle" the wheel as a "load". It will need more current at the start up with the load because it need a stronger rotational force to "move" the object from rest. Eventually it will speed up and speed up and becoming less of a load, then the C-EMF would generally reduce its motor input current and then the only thing needed between both RPM & Torque is really the higher RPM and low torque since the wheel is already in rotational motion. Rotational inertia anyone? :p

Please correct me if I'm wrong just some idea's to share.

Dre,
 
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BobK

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The executive summary: Less magnetic field means less back EMF, means more current drawn at the same voltage means more power drawn means motor spins faster.

Bob
 

Dretron

Jun 9, 2012
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The executive summary: Less magnetic field means less back EMF, means more current drawn at the same voltage means more power drawn means motor spins faster.

Bob

Exactly, or viceversa. The stronger the mag.field the greater the C-EMF the less current is drawn from the power source and he motor spins faster ;).
 

john monks

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Yes. The greater the C-EMF the less current is drawn for a given rotational speed.

Let's put it another way. For a perfect motor the C-EMF is exactly the same as your supply voltage and you will draw no current. If by magic you do nothing but increase the rotational speed your motor will turn into a generator and your current will be going in the reverse of the normal. You will be putting a charge back into the battery.

You are on the right track.
 

Dretron

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Yes. The greater the C-EMF the less current is drawn for a given rotational speed.

Let's put it another way. For a perfect motor the C-EMF is exactly the same as your supply voltage and you will draw no current. If by magic you do nothing but increase the rotational speed your motor will turn into a generator and your current will be going in the reverse of the normal. You will be putting a charge back into the battery.

You are on the right track.

Thanks. Good to know that point :).

Please tell me what do you think of this example is well fit:

"Now when the electric motor is turned on without the wheel attached as a load, it will instantly run easily and go fast because there is no mechanical resistance of the load. So the motor would eventually speed up and not need more current because of the C-EMF. We can agree that the hardest part was only the beginning.

Now when I attach a load. That is the wheel... The motor starts to draw more and more current for torque! And the RPM significantly drops and the C-EMF would be less generated.However, if the "LOAD" was not a major "resistance" to the motor... In a sense the motor can "handle" the wheel as a "load". It will need more current at the start up with the load because it need a stronger rotational force to "move" the object from rest. Eventually it will speed up and speed up and becoming less of a load, then the C-EMF would generally reduce its motor input current and then the only thing needed between both RPM & Torque is really the higher RPM and low torque since the wheel is already in rotational motion. Rotational inertia"

Please correct me if I'm wrong.
 

CDRIVE

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Yes, I think I know where this track eventually goes. All aboard for the O/U Express! :rolleyes:
 

Dretron

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Everything you said so far looks good to me.

Thats good to hear :)
Thanks John!

Yes, I think I know where this track eventually goes. All aboard for the O/U Express! :rolleyes:


What?

Over-unity... express? I was asking a simple question into relation of both TORQUE AND RPM and the effects of C-EMF in his whole process... I DID NOT UNDERSTAND how things worked. I could use what ever analogy to clarify my understanding...

Now I can make up a story about alien creatures to under more about space but that does not mean its true does it?

Anyways... I hope you don't assume things without asking questions and what is this leading to... I've just wirrten this from the top of my head not involving all the physical laws just a simple example.

To clear out things Im studying motor-efficiency. So this is relevant to it.
 

john monks

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I'm not sure what your asking.

I can tell you that nobody knows exactly what a magnetic field is, or an electric field is, or what gravity is. We know they exist because we see the effects of them. But when someone figures this stuff out someone is going to be rich. Someone is going to win the Nobel peace prize in physics. I tried to do this but I failed. Every other person who tried to explain this has failed. I hope to see someone do this before I die. I earned a degree in physics trying to figure this stuff out. But there are lessons I learned the hard way. These are:
1. Only study electronics from the physics, that is the electronic principles, and definitions.
We remember from experiments we conduct. We relate with fellow students to become fully versed in the language of electronics such as electrical terms, including units.
2. If you cannot derive a formula do not use it.
One professor taught me to solve all the problems using only variables and putting in the numbers only at the very last. This makes deriving formulas much faster and less likely to get lost in the math.

But I don't need to tell you how to study, you're already doing better than most people.
You're experimenting.
 
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