SIDE FURLING

[Breezee]

I have a 6 bladed rotor that is 8' in dia. Want to put a gravity tail vane on it, but was wondering if the offset on the rotor would be different than a 3 bladed rotor? Does someone know where I can find a formula or some info on size and weight of the tail? Also how far back from the rotor should the tail be, would the tail be offset from the yaw on the same side of the rotor or the opposite?

[hiker]

its all about size- prop speed-when you whant it to kick out--wind speed

 a 6 bladed prop u would think would have more backpressure than a 3 bladed prop

of the same size..[more blades that catch the wind]=more back pressure..

i know my little 7ft. single blade prop has a lot of back pressure in a good wind..

held it up in the air on day when there was a good wind--dam near knock me over..

just my thoughts............

[LEXX]

just a question, how do you have a single bladed prop or did you mean 2 bladed.

I'm confused!

[windstuffnow]

 Here is the basics for the tail weight...

Input:

  Rotor offset (meters)

  Diameter of Rotor (meters)

  Wind speed ( meters/sec)

  Tail tip weight (kg)

  Length of tail (meters)

Calculations:

  Rotor thrust = Diameter of Rotor^2 x windspeed^2/24

  Tail moment = Tail tip weight x Length of tail

  Rotor moment = Rotor thrust x Rotor offset

  Tail mount angle = sin-1(Rotor moment/Tail moment)

To find tail weight at a given tail mount angle:

  Tail moment = Rotor moment/ ( sin(angle in degrees)

  Tail weight = Tail moment/ Length of tail

Good to go!

Have Fun

Windstuff Ed

[Ungrounded Lightning Rod]

I have a 6 bladed rotor that is 8' in dia. Want to put a gravity tail vane on it, but was wondering if the offset on the rotor would be different than a 3 bladed rotor?

If  the rotor is efficient the number of blades doesn't matter.  (To be efficient, the lower the number of blades the higher the tip speed ratio.  Result is that, though there are fewer blades, they come around more often.  So the power pulled from the wind by the entire rotor, and thus the resulting drag for the entire rotor, is about the same - proportional to its area, not its number of blades.)

[Ungrounded Lightning Rod]

To be efficient, the lower the number of blades the higher the tip speed ratio.

Actually that's oversimplified.  Fewer blades can also gain efficiency by being wider or deeper.

The trick is to somehow see that all the air through the swept area gets decellerated by a given amount.  If you do that it doesn't matter how many blades you used to do it or what their shape is.  The drag (and the available power) is the same.

The Betz limit tells you how much to decellerate it to get the maximum power from it.  (Decellerate it too much and enough goes around the mill that you lose more by getting less air through the blades than you gain by pulling more power from the air that goes through.)

[Breezee]

Thanks for the info ED, could you try to put it in idot terms for me. I just barely made it out of high school.

[windstuffnow]

  Hello Breezee,

    All the variable interact with each other.  Suppose your rotor shaft is offset by 125mm, and, the thrust for furling is 40kg.  Then the moment is 40 * .125 or 5kg/m

    Find the moment of weight of the tail by weighing the tip.  If the tip weighs 10kg and the tail is 2 meters long then you have a 20kg moment of weight.

    The angle determines how much of that weight will be opposing the furling.  Furling moment is moment of weight times sine(angle).  If the moment of weight of the tail is 20kg/m and you want 5kg/m then the angle should be 14.5 degrees ( sine(5/20) ). My calculator requires the sin-1 (5/20) for that to come out in degrees.

Hope that helps

Windstuff Ed

[Breezee]

No wonder your in the wind power business!  Thanks ED

[windstuffnow]

  You can also use feet and lbs to come out with ft/lbs for the moments.  To find the actual thrust in lbs at a given windspeed....

.00492 x area of prop (in sq feet) x windspeed (in mph)^2 = pressure in lbs.

Have Fun

Windstuff Ed