1. Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

2.

3. Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?

ETA: By "attachment power" do you mean that the magnet can hold onto a 4kg mass against the force of gravity?

4. Originally Posted by Coldy
If this information is not enough for that kind of calculation - what else is needed?
You'll be using Faraday's and Maxwell's laws. You will need to know B-field, rate of change of that field (related to speed you move the magnet) the area of the receive coil and the number of turns. That will give you voltage. Power is a lot more complicated since receiver parasitics come into play.

5. Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Good questions already from m'learned friends. But I think it would also be good if you ca explain what form of linear motion you intend to use, to generate the power. Will this be a regenerative brake on a rail vehicle, or something? But if so, where does this permanent magnet fit into it? Or do you intend to harness some kind of reciprocating motion, and if so, what?

6. Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?
Ooh er, "magnet" alert. Let's hope this is not perpetual motion………….

7. Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?

ETA: By "attachment power" do you mean that the magnet can hold onto a 4kg mass against the force of gravity?
Actualy the magnet will induce electricity in the coil.
Yes, by "attachment power" I mean that the magnet can hold onto a 4kg mass against the force of gravity.

Originally Posted by billvon
Originally Posted by Coldy
If this information is not enough for that kind of calculation - what else is needed?
You'll be using Faraday's and Maxwell's laws. You will need to know B-field, rate of change of that field (related to speed you move the magnet) the area of the receive coil and the number of turns. That will give you voltage. Power is a lot more complicated since receiver parasitics come into play.
Well I'm not so competent in this feld so if you can give me some formulae for an example calculation?
There is some more data for this magnet:
1. Maximum energy product (B.H)max: 38 MGOe or 302 kJ/m^3;
2. Br = 12.6 kG or 1.26 T;
3. bHc = 11.2 kOe or 891 kA/m;
4. jHc = 12 kOe or 955 kA/m.
And lets say that the turns are 10.

Originally Posted by exchemist
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?
Ooh er, "magnet" alert. Let's hope this is not perpetual motion………….
Nope no perpetual motion here

8. Originally Posted by Coldy
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?

ETA: By "attachment power" do you mean that the magnet can hold onto a 4kg mass against the force of gravity?
Actualy the magnet will induce electricity in the coil.
Yes, by "attachment power" I mean that the magnet can hold onto a 4kg mass against the force of gravity.

Originally Posted by billvon
Originally Posted by Coldy
If this information is not enough for that kind of calculation - what else is needed?
You'll be using Faraday's and Maxwell's laws. You will need to know B-field, rate of change of that field (related to speed you move the magnet) the area of the receive coil and the number of turns. That will give you voltage. Power is a lot more complicated since receiver parasitics come into play.
Well I'm not so competent in this feld so if you can give me some formulae for an example calculation?
There is some more data for this magnet:
1. Maximum energy product (B.H)max: 38 MGOe or 302 kJ/m^3;
2. Br = 12.6 kG or 1.26 T;
3. bHc = 11.2 kOe or 891 kA/m;
4. jHc = 12 kOe or 955 kA/m.
And lets say that the turns are 10.

Originally Posted by exchemist
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?
Ooh er, "magnet" alert. Let's hope this is not perpetual motion………….
Nope no perpetual motion here
OK but what will make the magnet move and thereby provide the source of power for the generator to turn into electricity? You know, the equivalent of the turbine or the engine in a conventional rotary generator.

You do realise, do you, that you can't just get a flow of energy out of a static permanent magnet? A current is induced in a conductor that cuts magnetic flux lines. No cutting, no current.

9. Originally Posted by exchemist
Originally Posted by Coldy
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?

ETA: By "attachment power" do you mean that the magnet can hold onto a 4kg mass against the force of gravity?
Actualy the magnet will induce electricity in the coil.
Yes, by "attachment power" I mean that the magnet can hold onto a 4kg mass against the force of gravity.

Originally Posted by billvon
Originally Posted by Coldy
If this information is not enough for that kind of calculation - what else is needed?
You'll be using Faraday's and Maxwell's laws. You will need to know B-field, rate of change of that field (related to speed you move the magnet) the area of the receive coil and the number of turns. That will give you voltage. Power is a lot more complicated since receiver parasitics come into play.
Well I'm not so competent in this feld so if you can give me some formulae for an example calculation?
There is some more data for this magnet:
1. Maximum energy product (B.H)max: 38 MGOe or 302 kJ/m^3;
2. Br = 12.6 kG or 1.26 T;
3. bHc = 11.2 kOe or 891 kA/m;
4. jHc = 12 kOe or 955 kA/m.
And lets say that the turns are 10.

Originally Posted by exchemist
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?
Ooh er, "magnet" alert. Let's hope this is not perpetual motion………….
Nope no perpetual motion here
OK but what will make the magnet move and thereby provide the source of power for the generator to turn into electricity? You know, the equivalent of the turbine or the engine in a conventional rotary generator.

You do realise, do you, that you can't just get a flow of energy out of a static permanent magnet? A current is induced in a conductor that cuts magnetic flux lines. No cutting, no current.
Ops, I forgot to tell you that the magnet is moving trough the coil twice per second.
I dont want to uncover the source of the magnet movement before I make the real test. But if it works I can share the secret .

10. Originally Posted by Coldy
Originally Posted by exchemist
Originally Posted by Coldy
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?

ETA: By "attachment power" do you mean that the magnet can hold onto a 4kg mass against the force of gravity?
Actualy the magnet will induce electricity in the coil.
Yes, by "attachment power" I mean that the magnet can hold onto a 4kg mass against the force of gravity.

Originally Posted by billvon
Originally Posted by Coldy
If this information is not enough for that kind of calculation - what else is needed?
You'll be using Faraday's and Maxwell's laws. You will need to know B-field, rate of change of that field (related to speed you move the magnet) the area of the receive coil and the number of turns. That will give you voltage. Power is a lot more complicated since receiver parasitics come into play.
Well I'm not so competent in this feld so if you can give me some formulae for an example calculation?
There is some more data for this magnet:
1. Maximum energy product (B.H)max: 38 MGOe or 302 kJ/m^3;
2. Br = 12.6 kG or 1.26 T;
3. bHc = 11.2 kOe or 891 kA/m;
4. jHc = 12 kOe or 955 kA/m.
And lets say that the turns are 10.

Originally Posted by exchemist
Originally Posted by tk421
Originally Posted by Coldy
Hello,
I want to make myself a small linear alternator (generator) with NdFeB magnet (cilinder with dimentions: diameter - 10mm and 10mm height; and attachment power 4kg) and suitable coil. What power I can get from this and what coil will be best to use?
If this information is not enough for that kind of calculation - what else is needed?

Since the power doesn't come from the magnets, you have to let us know where the power is going to come from. I am not familiar with the term "attachment power," nor why it would have the units of mass (unless you are doing a mass to energy conversion with nuclear processes, which doesn't seem likely). Could you be a bit more specific about what is going to drive this thing?
Ooh er, "magnet" alert. Let's hope this is not perpetual motion………….
Nope no perpetual motion here
OK but what will make the magnet move and thereby provide the source of power for the generator to turn into electricity? You know, the equivalent of the turbine or the engine in a conventional rotary generator.

You do realise, do you, that you can't just get a flow of energy out of a static permanent magnet? A current is induced in a conductor that cuts magnetic flux lines. No cutting, no current.
Ops, I forgot to tell you that the magnet is moving trough the coil twice per second.
I dont want to uncover the source of the magnet movement before I make the real test. But if it works I can share the secret .
OK, I don't want to pry into your invention and please excuse me if I seemed patronising just now. It's just that we do get a, shall I say, variable standard of scientific understanding on this forum, so it's as well to check what new posters know fairly early on.

But it was your question about power output that bothered me and others. This will be a function of your power input, of course, from the power available in your source of reciprocating motion. But I suppose the trick will be to get a voltage out that makes it easy to take off in useable form the level of power your input can provide, i.e. a suitable amperage at a suitable voltage.

11. Originally Posted by Coldy
Ops, I forgot to tell you that the magnet is moving trough the coil twice per second.
I dont want to uncover the source of the magnet movement before I make the real test. But if it works I can share the secret .

I fear that you are chasing another "infinite energy" idea here, because you still seem not to acknowledge the "power input" part of the problem, as exchemist has pointed out. In the boldfaced text above is a key phrase. What is making that magnet move? As you've stated the problem so far, I could get any amount of energy out of the device. And I don't think that you know why.

A common misconception is that a magnet moves through or past a coil with no drag. That misconception leads to "free energy" fantasies. In actual fact, a magnet experiences something akin to viscous drag, with that drag increasing as one attempts to pull more power out of the coil. Despite there being no apparent physical connection between the magnet and coil, there is an electromagnetic field that mediates the transfer of energy from the magnet to the coil (and vice-versa). You push the magnet, and you ultimately push the electrons in the coil. The more energy you try to deliver from the coil, the harder you must push the magnet. You can't just set up a magnet on a pendulum, say, and expect it to push freely through a coil twice per second, even if you eliminate air drag and have perfectly lubricated pivots. As soon as you put a load on the coil, the pendulum will slow, then stop.

So, when you say "the magnet is moving through the coil twice per second," you've actually missed the point entirely. I'll say it again, for you seem to have continued to miss its signficance: The power comes from whatever is making the magnet move, not from the magnet itself. And that power will grunt harder the more energy you try to take out of the coil.

"No free lunch" at work again.

12. Originally Posted by Coldy
Ops, I forgot to tell you that the magnet is moving trough the coil twice per second.
I dont want to uncover the source of the magnet movement before I make the real test. But if it works I can share the secret .
OK good now we're getting somewhere. We have frequency, which is critical since EMF is proportional to the derivative of the B-field. We have area, number of turns and total B field. With this you can calculate the EMF (voltage) that the generator will produce, which is the first half of your answer. (Not going to do this for you, but google "electromagnetic induction" and the equations are right there.)

The second half of your answer will be current. Any current drawn from the system will cause an opposing force on the magnet, so the prime mover is going to become important. Your load will determine how much current will be drawn so that's also important.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement