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A little help with some math please...?

N

Nick.

Jan 1, 1970
0
This is something I've been scratching my head over for a couple of
days...

I want to model a permanant-magnet type (bicycle) AC generator, to
calculate its output at a given speed / load. Characteristics of the
actual device can be seen here:

http://www.nabendynamo.de/12vu.gif (the plots are no-load and two
different load values)

I know that the generator has a finite current limit, due to the induced
current generating its own magnetic field (someone correct me if I'm
wrong), and I can *see* what is occurring, but I just can't seem to define
it mathmatically :eek:(

I need to calculate Vout as a function of rotational speed and load.

This is frustrating because it's pretty basic, but hopefully is one of
those cases of just getting past the starting post. Before anyone asks, no
this is not a homework project (FAR to many years since I was at college)
nor anything further than hobbyist tinkering.

Thanks,
Nick
 
J

John Popelish

Jan 1, 1970
0
Nick. said:
This is something I've been scratching my head over for a couple of
days...

I want to model a permanant-magnet type (bicycle) AC generator, to
calculate its output at a given speed / load. Characteristics of the
actual device can be seen here:

http://www.nabendynamo.de/12vu.gif (the plots are no-load and two
different load values)

I know that the generator has a finite current limit, due to the induced
current generating its own magnetic field (someone correct me if I'm
wrong), and I can *see* what is occurring, but I just can't seem to define
it mathmatically :eek:(

I need to calculate Vout as a function of rotational speed and load.

This is frustrating because it's pretty basic, but hopefully is one of
those cases of just getting past the starting post. Before anyone asks, no
this is not a homework project (FAR to many years since I was at college)
nor anything further than hobbyist tinkering.

Thanks,
Nick

I think a reasonable model for the generator is an ideal AC voltage
source with voltage proportional to frequency (essentially the
unloaded curve)) in series with a resistor (winding resistance) and an
inductor. The inductor is selected to produce the drooping light load
curve, and the resistor is selected to tweek the fit on the high load.
 
I

Ian Bell

Jan 1, 1970
0
Nick. said:
This is something I've been scratching my head over for a couple of
days...

I want to model a permanant-magnet type (bicycle) AC generator, to
calculate its output at a given speed / load. Characteristics of the
actual device can be seen here:

http://www.nabendynamo.de/12vu.gif (the plots are no-load and two
different load values)

I know that the generator has a finite current limit, due to the induced
current generating its own magnetic field (someone correct me if I'm
wrong), and I can *see* what is occurring, but I just can't seem to define
it mathmatically :eek:(

I need to calculate Vout as a function of rotational speed and load.

This is basically a permanent magnet dc motor operating as a generator. The
equation for a pm dc motor is:

Vt = IR -kw

Where Vt is the terminal voltage, I the motor current, R the winding
resistance, k a constant and w the rotational speed. The kw term is the
back emf generated by the motor turning.

I am sure this will do for you with suitable alteration of the terms.

HTH

Ian
 
R

Rich Grise

Jan 1, 1970
0
This is something I've been scratching my head over for a couple of
days...

I want to model a permanant-magnet type (bicycle) AC generator, to
calculate its output at a given speed / load. Characteristics of the
actual device can be seen here:

http://www.nabendynamo.de/12vu.gif (the plots are no-load and two
different load values)

I know that the generator has a finite current limit, due to the induced
current generating its own magnetic field (someone correct me if I'm
wrong), and I can *see* what is occurring, but I just can't seem to define
it mathmatically :eek:(

I need to calculate Vout as a function of rotational speed and load.

Just do a curve-fit - you've got it right there:
Effektivspannung In Volts per Fahrgeschweindigkelt In km/h. ;-P

It looks kinda like a log, but heck, you should be able to curve-fit
something like that in only order 2. :)

Good Luck!
Rich
 
T

Ted Edwards

Jan 1, 1970
0
John said:
I think a reasonable model for the generator is an ideal AC voltage
source with voltage proportional to frequency (essentially the unloaded
curve)) in series with a resistor (winding resistance) and an inductor.
The inductor is selected to produce the drooping light load curve, and
the resistor is selected to tweek the fit on the high load.

The inductor is probably the most important. Since frequency is
proportional to rpm and inductive reactance is proportional to
frequency, this makes the voltage self limiting at the design load.

Ted
 
M

Marte Schwarz

Jan 1, 1970
0
Hi Nick,
I want to model a permanant-magnet type (bicycle) AC generator, to
calculate its output at a given speed / load. Characteristics of the
actual device can be seen here:

http://www.nabendynamo.de/12vu.gif (the plots are no-load and two
different load values)

Ein zusätzliches Problem bei der Beschreibung dieses Systems ist die extrem
nichtlineare Kennlinie der Leuchtmittel. Die sind es nämlich IMHO die den
Gleichstromcharakter des ganzen Systemes einprägen. Glühobst neigt nämlich
zu einer Art Stromkonstanter zu werden. Ich hab zu Hause noch ein Ladegert,
das genau so seinen Strom einstellt und das gar nicht allzu schlecht. Sei
doch mal so nett und mess das ganze für uns mal mit ohmschen Widerständen
als Last aus.

Sorry for all english readers, german is much easer for me ;-)

Marte
 
J

John Popelish

Jan 1, 1970
0
Ted said:
The inductor is probably the most important. Since frequency is
proportional to rpm and inductive reactance is proportional to
frequency, this makes the voltage self limiting at the design load.

Agreed. The value of resistance that best fits the curves may
approach zero.
 
N

Nick.

Jan 1, 1970
0
Ein zusätzliches Problem bei der Beschreibung dieses Systems ist die
extrem nichtlineare Kennlinie der Leuchtmittel. Die sind es nämlich IMHO
die den Gleichstromcharakter des ganzen Systemes einprägen. Glühobst
neigt nämlich zu einer Art Stromkonstanter zu werden. Ich hab zu Hause
noch ein Ladegert, das genau so seinen Strom einstellt und das gar nicht
allzu schlecht. Sei doch mal so nett und mess das ganze für uns mal mit
ohmschen Widerständen als Last aus.

Hi Marte

No worries... Google Translate sort of did the job :eek:) I'm not too sure
what Glühobst means (Google translates to Glow Fruit)

As well as the graph mentioned, I also have some measured data, using
resistive loads. (I suspect that the graph might've been measured using
resistors) Good point, though.

Nick
 
N

Nick.

Jan 1, 1970
0
I think a reasonable model for the generator is an ideal AC voltage source
with voltage proportional to frequency (essentially the unloaded curve))
in series with a resistor (winding resistance) and an inductor. The
inductor is selected to produce the drooping light load curve, and the
resistor is selected to tweek the fit on the high load.

I don't think that'll work John, because if you also plot (for a given
speed/freq) voltage vs load, you will get an exponential-ish curve (up to
the no-load voltage)
 
N

Nick.

Jan 1, 1970
0
Just do a curve-fit - you've got it right there: Effektivspannung In
Volts per Fahrgeschweindigkelt In km/h. ;-P

It looks kinda like a log, but heck, you should be able to curve-fit
something like that in only order 2. :)

Good Luck!
Rich


Thanks, I'll probably end up doing that. But.. one of the points of the
exercise is to try to understand the principles behind what effectively is
a few turns of wire wrapped around a revolving magnet. (20yrs ago this
would've been fairly trivial, but I'm afraid I then hung up my soldering
iron for a keyboard. Scary how much one forgets...)

On the practical side, all I want to do is vary the load (power a bulb,
maybe two, charge a battery) based on speed. I can do that easily enough
using measured data in a lookup table.

Cheers,
Nick
 
J

John Popelish

Jan 1, 1970
0
Nick. said:
I don't think that'll work John, because if you also plot (for a given
speed/freq) voltage vs load, you will get an exponential-ish curve (up to
the no-load voltage)

If the voltage source produces a voltage and frequency proportional to
speed (what the graph shows for no load, for voltage, anyway), and the
inductor has an impedance proportional to speed (since the generator
frequency is proportional to bike speed), you get a constant current
into any load resistance that is significantly lower than the inductor
impedance. That point occurs for a different speed for each load
resistance, at a lower speed for lower load resistances, and that is
generally what the graph shows. No exponentials anywhere in the AC
analysis.
 
M

Marte Schwarz

Jan 1, 1970
0
Hi Nick,
No worries... Google Translate sort of did the job :eek:) I'm not too sure
what Glühobst means (Google translates to Glow Fruit)

Glühobst is a more general description for glowing things not only "-birnen"
;-)
A "Quartzstrahler" (is it called glass-heating-cylinder?) is a nice example
with a similar behaviour.
As well as the graph mentioned, I also have some measured data, using
resistive loads. (I suspect that the graph might've been measured using
resistors) Good point, though.


Well in the graph there was the explanation with 3W and 2.4 Watt lamps
AFAIR.
The differences in the graphs would be very interesting for me too. I use
such a nabendynamo in my bike and want to design my accu-frontlight to use
the daily energy as a standby-light in the frontlight too. I have a old bisy
but it should be named buggy-light ;-)

Marte
 
R

redbelly

Jan 1, 1970
0
I'm getting a pretty good fit using the following:
(view in fixed-width font, eg. Courier)

1
Vout = ------------------------------------
SquareRoot[ (A/speed)^2 + B^2 ]


Where the parameters are

A & B = 1.1 & 0 for the no-load case
A & B = 1.25 & 1/17.7 for the middle curve
A & B = 1.4 & 1/7.9 for the lowest curve

Note: at low speed, Vout is approx = speed / A
at very high speed, Vout is approx = 1 / B

Regards,

Mark

p.s. these parameters are not optimized least-squares values. I just
overlayed your .gif file on an Excel graph and played around till
things looked reasonable.
 
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