# How does this Snowmobile Generator work?

#### bdubz

Joined Sep 24, 2021
6
Hi folks, I am a little bit confused on the conversion of torque to electrical power. specifically i have a rotor stator combo I've been looking at from a snowmobile (photo below) and I simply do not understand it. I know it has multiple different turn ratios on the bobbins for different purposes which they generate higher voltages. About 6 of the coils are called the lighting coil which generate the power for the 12 V system on the machine. What i dont understand is how the torque load on the engine will ever change with this type of configuration. As the rotor (with its permanent magnets) passes by each of the coils it generates a voltage depending on the strength and the speed of the magnetic field dictates the amount of current i think... Either way regardless of how much electrical load is "requested" outside the stator, the stator should impose the same torque on the rotor or so i thought. when i load down the generator the whole system makes a chattering noise which tells me the electrical load is created a larger torque load. Can someone explain how this works and show the equations to how a system like this might function?

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#### ronsimpson

Joined Oct 7, 2019
2,540
I have a small tractor that has a generator like that. (Bigger!)
the stator should impose the same torque on the rotor or so i thought
I don't think so.

#### LowQCab

Joined Nov 6, 2012
2,651
A "chattering-noise" is NOT normal.
Find out what's coming loose before it causes serious damage.
Knowing how it works Electrically is not going to solve a Mechanical problem.
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#### crutschow

Joined Mar 14, 2008
31,138
As the rotor (with its permanent magnets) passes by each of the coils it generates a voltage depending on the strength and the speed of the magnetic field dictates the amount of current i think... Either way regardless of how much electrical load is "requested" outside the stator, the stator should impose the same torque on the rotor or so i thought.
Nope. That would mean the output electrical power is independent of the input mechanical power.

Actually, the force required for the magnet to move past the stator windings is proportional to the load current through the windings.
The winding current generates a magnetic field proportional to the current, of polarity that opposes the movement of the rotor magnet.
That's how the energy is transferred from the rotor rotation through the magnetic field to the electrical power output.

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#### BobTPH

Joined Jun 5, 2013
6,086
Current flowing in coils create a magnetic fiekd that pushes against the magnets in the rotor. When the lights are on, current flows and high torque, when off, no current and low torque.

#### Janis59

Joined Aug 21, 2017
1,537
Voltage is dependant on bobbin turn count and SPEED of magnet, and, of course, the magnet Teslas value.
Force is depended on the same said but the Amperage.

Let say, the electrical output is about 80-90% of input mechanical power (energy conservation law). Then V(out)*i(out)=0.9*F(rot)*v(rot) at radius of bobbin centers.

All the formulas may find in textbook Physics for form 9 and better level at Physics for form 11 (1Form 12 tells about high frequency and nuclear physics). At English may find to worth reading the Giancoli, Collegiate Physics, International edition. Or Hyperphysics.com. Or Engineeringtoolbox.com. And no doubt, Wikipedia.

#### bdubz

Joined Sep 24, 2021
6
A "chattering-noise" is NOT normal.
Find out what's coming loose before it causes serious damage.
Knowing how it works Electrically is not going to solve a Mechanical problem.
.
.
.
well the issue is all the snowmobiles that use this generator make that chattering noise at very high load or so I've heard. i know its obviously not a good thing but the stator is bolted down solid and the rotor is press fit on the crankshaft which spins with the engine so i don't even know where the source of the noise is!

#### bdubz

Joined Sep 24, 2021
6
Current flowing in coils create a magnetic fiekd that pushes against the magnets in the rotor. When the lights are on, current flows and high torque, when off, no current and low torque.
Thanks for the info! Do you have an equation or source material i could read to understand how the coils magnetic field interacts with the magnets? That's really the goal of this for me is to understand the torque imposed on the rotor based on the electrical load on the vehicle system. I.E. if i turn on some headlights, how much more torque will be applied on the rotor and in turn the engine. I know with Alternators this can be adjusted by changing the current in the electromagnets to create stronger magnetic fields but in this case the magnets are permanent so I'm not understanding how its changed...

#### BobTPH

Joined Jun 5, 2013
6,086
The additional torque is such that the power required to turn it is equal to the electric power delivered.

The electrical power is V x I. Don’t know the formula for power required to turn a shaft based on torque off the top of my head.

#### BobTPH

Joined Jun 5, 2013
6,086
Okay, I looked it uo.

To rotate a shaft with torque T, at n RPM the power in Watts is:

P = 2π T n / 60 = V I

T = (60 V I ) / (2 π n)

#### LowQCab

Joined Nov 6, 2012
2,651
Horsepower is directly convertible to to Kilowatts, ( thousands of Watts ).
Calculate how many Watts You expect to draw from the Alternator,
multiply by an unknown Efficiency-Loss-Factor,
then divide the derived number by 1000,
this number in Kilowatts can then be converted to Horsepower.

1 Horsepower = 0.74 Kilowatts
1 Kilowatt = 1.34 Horsepower

But Horsepower it's self is the result of a Math-Formula ........
Horsepower is .........
Pounds/Feet of Torque X RPM / 5252 = Horsepower.

So for a given amount of Electrical-Load-Watts,
less and less Torque will be required as the RPM increases higher and higher,
or,
to look at it from the other-way-'round ..........
the Alternator will be capable of more Power-Output in Watts, as the RPMs increase.

50-Watts / 1000 = 0.050 Kilowatts,
0.050 Kilowatts X 1.34 = 0.067 Horsepower,
this is also equivalent to 0.067 Pounds/Feet of Torque at 5,252 RPM,
( note ...... this does not account for inefficiencies )
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#### LowQCab

Joined Nov 6, 2012
2,651
Voltage Regulation of a Permanent-Magnet-Alternator can be quite tricky to do EFFICIENTLY.
Most Motorcycles, and other small Engines,
just apply MORE-ELECTRICAL-LOAD than the Alternator can supply.
This is usually in the form of a "Shunt-Regulator", which turns the excessive Power into HEAT.

There are other factors, such as Inductive-Reactance,
which can also be employed to regulate the Output-Voltage.
This can easily get very complex, and can't be explained in just a few paragraphs.
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#### crutschow

Joined Mar 14, 2008
31,138
I know with Alternators this can be adjusted by changing the current in the electromagnets to create stronger magnetic fields but in this case the magnets are permanent so I'm not understanding how its changed...
What changes is the opposing field generated by the change in the stator load current.

#### bdubz

Joined Sep 24, 2021
6
What changes is the opposing field generated by the change in the stator load current.
ah i see. I think this was the information i was really missing. I was just thinking about the imposing magnetic field from the permanent magnets and not the opposing magnetic field generated by the current flowing through the coils. The more current you ask for, the more intense the field becomes and the harder it is to pass the permanent magnet through it.

#### bdubz

Joined Sep 24, 2021
6
Voltage Regulation of a Permanent-Magnet-Alternator can be quite tricky to do EFFICIENTLY.
Most Motorcycles, and other small Engines,
just apply MORE-ELECTRICAL-LOAD than the Alternator can supply.
This is usually in the form of a "Shunt-Regulator", which turns the excessive Power into HEAT.

There are other factors, such as Inductive-Reactance,
which can also be employed to regulate the Output-Voltage.
This can easily get very complex, and can't be explained in just a few paragraphs.
.
.
.
Right and i think this is what was really confusing me because I believe the regulator on this system is a shunt type so its always pulling a lot of power and burning off the rest so it never looked like the load changed. but really what changed was the amount of heat generated in the ?regulator? i guess is where it burns it off? or maybe recycles it back through the coils to burn off there. But either way with a shunt type regulator it seems like applying as much electrical load as you can actually helps the vehicle stay cool otherwise that power gets consumed as heat and the torque load stays the same...

#### crutschow

Joined Mar 14, 2008
31,138
My motorcycle had a shunt regulator that momentarily shorted the windings to ground using SCRs for each AC cycle when the voltage exceeded the desired value.
Thus the excess power was dissipated in the SCR forward voltage drop and the generator winding resistance.
The regulator got too hot to touch when the engine was running.

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#### BobTPH

Joined Jun 5, 2013
6,086
ah i see. I think this was the information i was really missing. I was just thinking about the imposing magnetic field from the permanent magnets and not the opposing magnetic field generated by the current flowing through the coils. The more current you ask for, the more intense the field becomes and the harder it is to pass the permanent magnet through it.
I guess you missed my post#5 then.

#### LowQCab

Joined Nov 6, 2012
2,651
"" .............. it seems like applying as much electrical load as you can actually helps the vehicle stay cool
otherwise that power gets consumed as heat and the torque load stays the same ......... ""

??????????????
"vehicle stays cool" ????, what part of the Vehicle ?

"power gets consumed as heat"
( Power may be converted into a different form, Heat is just one form )

People "consume" food,
Power can be "converted" into some other form, in this case, that's usually Electricity or Heat,
any Heat will raise the Temperature of various associated parts,
and eventually, will be "dissipated" into the surrounding, cooler, Air.

All of the components related to the Regulation of the Alternator will dissipate more Heat,
as the RPM of the Engine increases.
More Power is being produced at higher RPMs, and that Power must go somewhere.
The extra Power can not be allowed to go to the various Loads,
because the Loads have a definite limit as to how much
internal-Heat they can withstand before failure.
Therefore, the Power to the Loads must be "Regulated" within certain limits,
and the excess Power must be converted to Heat, instead of "Electromotive-Force", ( Voltage ).
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