In the posting http://www.fieldlines.com/story/2009/5/12/194822/794 (Scored a treadmill, questions about the motor) you state:

I show a 12 vdc cut-in at 320 rpm
and a 14.7 volt dump at about 400 rpm

a little fast but plenty workable . . .

Has that answered my question for me? eg I'm looking for 12v at XXXRPM = 10m/s?

Also, does that mean that if the motor is plated with a voltage and rpm the voltage at XXXrpm can be calculated? eg I can calculate XXXRPM for 12v from the motors running voltage and rpm?

Thanks

What you get out will depend on the generator efficiency and you can only measure that with some form of torque measurement at the chosen speed when charging a battery.

Either side of your chosen wind speed things will go wrong so how well the whole system performs depends how badly things go wrong as you depart from your calculated blade optimum. With a single point loading you should in theory hit the highest point on your prop Cp curve and the overall efficiency will be Cp x generator efficiency.

Either side of that chosen point it may be way off.

The usual method is to choose some minimum useful wind speed and fix cut in at that point. You then try to load the thing such that you maintain a reasonable prop Cp over the useful wind range ( and it may be to your advantage to get this optimum near your 10m/s). In higher winds your loading will not be effective but due to the cube law of wind power the output will still increase and you should have plenty of power in the high winds and you will almost certainly need to spill power in some way to prevent generator burn out. This method will give you the best energy capture, fixing an optimum point at one wind speed without considering performance over a wind speed range almost certainly will not.

If you have no means of measuring the efficiency of your generator you can only guess at it and the efficiency will be load dependent. At a guess something like an Amtek may reach 80% efficiency at a few watts and it will probably fall to about 50% at its nameplate rated current. If you drive it well beyond its nominal rating to get something in the hundreds of watts that some aim for then efficiency is likely below 30%. Depending on your chosen blade size at the chosen 10m/s you may be looking at efficiencies of 70% down to about 30 so no point in doing clever maths on the prop unless you can settle the generator efficiency.

Thanks for taking the time to post the answers.

Seems everytime I ask something I open a can of worms :-)

First thing I should say is that when I said 10m/s I actually meant 10mph or 4.5m/s (sorry it was late at night and I got my mph and m/s mixed up)

I'm still not sure what I'm aiming for?

Norm

You state:

maybe
yyyyy will be about 900 rpm will give you a 3amp
charging rate with a 30 volt Ametek.

Am I aiming for an amperage of 3amp or is that what you think the Ametek will produce?

Presumably its the output in amps that affects the charging of the battery?  But what about this figure of 18v that is stated as required to charge a 12v battery?

Flux/Norm

Thanks for pointing out that presumably there's no point in running any test without the generator connected to a battery?  Does it need to be the same battery as that which will be used in the setup, eg does a 120AH battery need more torque to charge than say an 80AH (I assume so since the resistance must be different)?

Flux

I'm really not that sure about anything much - it's pretty steep learning curve (but fun for all that).  I've been using Hugh Piggott's Windpower Workshop and that's where the 4.5m/s came from.  He mentions in more urban environments that 3m/s is nearer what's achievable - so I ideally that's what I was hoping for rather than 4.5m/s - not sure if that sort of wind speed will give me anything useful with the Ametek.

You state

"The usual method is to choose some minimum useful wind speed and fix cut in at that point."

Presumably the minimum useful wind speed would be lower than whatever your optimum speed (say 3m/s) is, since you wouldn't expect to get this speed all the time?  Also at cut in speed the system is producing about 12v which is not sufficient to charge the batteries.

"In higher winds your loading will not be effective but due to the cube law of wind power the output will still increase and you should have plenty of power in the high winds and you will almost certainly need to spill power in some way to prevent generator burn out."

One way to spill the power would be through the furling mechanism I presume?

The optimum speed will therefore fall somewhere between the cut in speed and the maximum speed at which you wish to "spill power".  Between these speeds whatever blade design/size etc needs to provide the Cp necessary to drive the generator?

I would want a bit of power before 10MPH.
I would try for blades in 7 or 7.5MPH wind to have the motor making about 13V open.

Thanks for the replies - You're right I am giving myself a bloody headache - I'm glad to hear it's unnecessary :-)

I think that's what I was trying to get a handle on - something simple like "try for 13v at 7mph (or 10mph)".  With that I can start experimenting.

That's great

Thanks for taking the time to sort my muddle out.

Regards

Again thanks for all the replies.  I now have the data I need to get started.  I'll let you know how I get on.

Again many thanks

Regards