Furling physics

[imsmooth]

The rotor creates a thrust, that when applied to the offset from the axis creates a moment.  When this thrust exceeds the moment of the tail due to gravity the windmill will start to furl.  I have no problems with this part.



This all assumes that the tail is fixed parallel with the wind.  Nowhere did I see any consideration to the area of the tail.  For an extreme case where the tail has zero surface area the tail will move with any rotor thrust because nothing is keeping the tail fixed in position.  The rotor thrust does not have to exceed the weight of the tail because the wind can't keep it in position.



So, should the area of the tail at least equal the area of the blades?  Does anyone have any answers or thoughts?

[wpowokal]

Have you read this post.

http://www.fieldlines.com/story/2005/3/9/232440/9997

[imsmooth]

I have seen it and it explains furling, but it does not answer my question.  Point #3 states that the tail needs to be large enough to stay in the wind, but it does not mention any calculations to determine the critical surface area the tail must exceed.



Clearly, once the tail is large enough it will stay in the wind and not get pushed through it as the rotor turns.  But if the tail is too small as the rotor turns, and there is not enough wind to lift the weight of the tail on the angled hinge, it will get pushed through the wind.



So, does anyone want to take a crack at this?

[imsmooth]

I'm going to take a crack at this:



P = Power/sec = F*distance/sec = F*V, or P/V = F (thrust force)

Also, P per second = (0.5)*M*V^2, where the mass of air/second = M = r*A*V, where r = 1.225 kg/m^3


Therefore, F = P/V = 0.5 * r*A*V * V^2 / V


F = 0.5*(1.225)*A*V^2



However, the wind initially is parallel to the tail so there is no component striking it.  If the tail moves 1 degree the horizontal wind component is V*sin(1).  Once the tail moves about 5 degrees the component is V*sin(5)



The rotor thrust, which we must balance, is D^2 * V1^2 /24



Thus, 0.61 * A * V^2 * sin(x) = D^2 * V1^2 / 24.  The velocity drops out giving:

A = D^2 / 14.7*sin(x)



For an 8 foot rotor, D = 2.64m and when the tail is 1 degree into the wind we need a tail area of 27m^2 to not get pushed by the rotor thrust.  This is a little big.  Once the tail has moved 5 degrees into the wind the air is getting a better hit on it and the necessary area to stay at this point is A = 5.4m^2



This makes sense.  When the tail is parallel with the wind nothing is holding it there and it will get pushed by rotor thrust.  Once it has moved over a little the area should catch enough wind so that it is held in place until the thrust exceeds the weight of the tail and lifts it.  However, the tail stays close to parallel with the wind.  If the tail is too small it will get pushed through the wind.  The equation also makes sense in that as thrust increases with wind velocity so does the thrust on the tail.



Any comments?  

[BruceDownunder]

 Hey,,come on ,,, you saying a 8 foot dia mill needs a tail on it of 5.something square metres... it will NEVER  furl ...

forget the theory,,just do what us guys do and follow Hugh or filedlines or the posts on furling from this forum --

I used 1/8 inch marine plywood,around 1.5 metre square,,added some weight till I liked what I say when the wind blew strong,,works perfect,,,been that way for 3 years-plywood is still perfect. shove the plywood tail section out on a boom around the length of a blade or a wee bit more..

best of luck

Bruce

[imsmooth]

I don't know if this theory is correct.  Also, don't confuse square area with weight of the tail.  The tail can weigh less than a kilo but still have a large area.  The area I am referring to simply relates how easily that tail wants to stay parallel with the wind.  As the rotor turns the tail move up along the slanted axis, but the tail is still close to in-line with the wind.  The tail stays with the wind; it is the axis and rotor that turn.  If the tail is too small it just gets pushed around the axis.  This was my question.



With a 20 degree offset an 8' rotor will only need about 1 m^2.  I also didn't compensate for the fact that as the whole axis turns the thrust goes down, so a smaller tail area is needed.  I guess a cos(x) factor needs to multiply the rotor thrust.



I'm not saying I'm right, but I would like to hear what others have to say. Again, this is a theoretical question.

[imsmooth]

SOrry for the double post, but another thought.  Maybe the minimum area should just be equal to the area of the blades?

[BruceDownunder]

 I'd say the 1 Mtr2 tail fan would be about right ,,looking out the window at mine .. But you may still have to weigh it with around 3 Kilos .. I spent many hours watching and wondering what could be done ,finally got it just about rightby adding weight till it behaved itself..

Go for that on your 8 footer

Bruce

[wpowokal]

My point is there are many variables, angle, weight, area, offset, blades ability to seek the wind etc....

OK make it 10% of swept area as a starting point. My duel rotor has 3m diamiter blades, tail length 1.2m and area of 0.54msq. The only weight is in the ply itself and the steel of the tail.

It furles well to limit the unit to 1 Kw and in extreem gusts briefly rises to 1.5Kw.

allan down under

[Flux]

Jonathan your theory is fine but once again there are so many conflicting requirements that everything has to be a compromise.

The basic idea of furling starts off by assuming that the tail is anchored directly down wind. In real life it never can be without being far too big to be practical. You have to accept that it will run at some angle to the wind.

If you use a very light material then you can have a bigger vane and run at a small angle to the wind but enormous vanes are not without problems.

Most people seem to use plywood and that puts serious restrictions on the area you use as you need to keep it light enough to furl. Usually a compromise is made where a balance is met with the tail offset about 20 deg away from furling.

I am always reluctant to enter this sort of discussion as I never do this very scientifically, the size is usually determined by looks and what is lying around so I have little idea of the sizes I use. I do tend to err on the small side and usually have to offset at least 20 deg.

I can't imagine using anything more than 1m^2 for an 8ft machine, I would more than likely  keep it to to about 6ft^2 or even less. That would be about 16% of rotor area. I can't help feeling I have never used as big as this but it seems a reasonable starting point.

I also tend to use longer booms, for 8ft I would make it 4 ft to the front of the vane but some may make it 4ft to vane centre, again a personal opinion based largely in my case on what I think looks right.

The whole thing is a compromise and in turbulent areas it's not a very good one. Its simplicity is the saving feature but in terms of performance and easy life on a machine the waggling tail would be the first thing to go if I went to a pitch controlled hub.

Flux

[finnsawyer]

"P = Power/sec = F*distance/sec = F*V, or P/V = F (thrust force)"

I assume you really mean P is power.  Then P = time rate of doing work = joules/sec.

Basically, it is the rate at which a force changes the kinetic energy of a mass, M.  The kinetic energy of the mass is then = 0.5xMxV^2 and has the units of joules (MKS system).  The force is said to do work in that case.  The expression P = FxV is correct when F and V as vectors point in the same direction (a vector form takes care of the general case).  But the force has to be doing work.  It can't just be pushing against something that is not moving.  In the case when the wind is blowing steady and the tail and rotor point in fixed directions the forces are doing no work.  One then must balance forces right up until the wind speed is such that the turbine furls.  I hope this points you in the right direction.

By the way, did we have a similar discussion once before?

[imsmooth]

I understand there are a lot of variables.  I am just trying to develop some basic theory.  I also want to point out that the rotor thrust equation assumes ideal thrust with maximum power extraction at the Betz limit of 59%.  If the rotor is real-world and extracts, say 30%, the tail area will be half.



A = K*cos(x)*D^2 / 14.7 * sin(x), where x is the accepted tracking angle to the wind, and K is the percentage from the ideal rotor.  Using the example of an 8' (2.64m) rotor and an angle of 15 degrees, and a rotor with 25% efficiency (K = 25/59 = 0.42): A = 0.7 sq meters = 6 sq feet.  This seems right.  The tail will track the wind at about 15 degrees off.  Once the rotor thrust exceeds the weight vector of the tail momement based on its length and weight it will begin to furl, maintaining the 15 degree offset from the wind.



This is just theory. I hope it helps in some way.

[Flux]

Yes I know some of you like to reduce everything to nice theory, at least it is a starting point.

You are right that the thrust applies to maximum power extraction. I also think you are right that below betz limit the thrust is reduced.

What I don't think any simple theory can cope with is the effect of the prop seeking force that acts on the restoring couple in a different way from the thrust.

Your ideas imply that a stalled rotor will furl at a higher wind speed than a correctly loaded one. It may be true as we have little data on the actual furling wind speeds, most data is related to output amps and wind speed.

One thing seems certain, and that is that a machine that furls satisfactorily in stall mode may not furl at a safe speed( or may fail to furl at all) if got out of stall by some means.

Unless you can adapt your theory to account for this seeking force then it remains little more than a mathematical exercise.

I don't think this is a simple issue otherwise a lot of commercial manufacturers would not have fallen into lots of traps over the years. I still pass machines on a daily basis with tails set at the wrong angle and they start out half furled. I questioned this with the manufacturers and they assure me the machines are assembled correctly. Perhaps they never come here to see the discussion.

Flux

[scottsAI]

While digging around fieldlines for data on the relationship of Blade dia to tail area (see table below third column). Data not consistent, drawing a line through the points and eye balling the areas ratio, 16:1 is the best fit. The tail size for 10 ft was most consistently 5 sqft. Made sure line was near 5 to come up with the ratio.

Did not come across any definitive formula for the tail length.

Observations: Tail's length equal to the blade dia works.

Tail's turning torque = (tail area) * (length).

Half the area requires twice the tail length.

Weight of the tail is another factor we will cover later.

Blade ___tail 1/16

dia __area _rec area

(07) (38.4) (3) (2.4)

(08) (50.3) (5) (3.1)

(10) (78.5) (5) (4.9)

(12) (113) (12) (7.1)

(14) (153) (10) (9.6)

(16) (201) (10) (12.6)

Your Formula for the blade drag force looks good:

http://en.wikipedia.org/wiki/Drag_equation

Spinning blades behave as a disk blocking the wind (with holes), Cd adjust how much.

Without proof I would expect Cd should be a number below Benz limit.

We are NOT talking about the drag resistance of the blades spinning in the air as a lifting surface, we are talking about the drag force resulting from the wind blowing against the blades. Trying to bend the tower over.

Imsmooth your equations are missing the blades offset distance times the (wind) force producing the Furling torque which the tail must counter to keep the blades normally pointed into the wind. The tail torque (tail Weight as force)*(effective length of the tail) at the sloped hinged counters the blades furling torque. http://en.wikipedia.org/wiki/Torque

Effective length of the tail - is the distance from its center of gravity, NOT its length.

The torque vectors must include the distance the hinge is from tower center.

Unfortunately these vectors are difficult to simulate, I can't give you exact answers... yet.

DanB post, put there by wpowokal at the beginning of this post:

http://www.fieldlines.com/story/2005/3/9/232440/9997

Very helpful in understanding. Thanks DanB.

Furling

Once the furling force exceeds the tail weight the blades will turn from the wind. Turned blades reduce the effective area of the blades to the wind reducing furling torque. The furling torque is a squared function as compared to the sin(blade angle) function. I expect once the furling wind speed is reached it takes but a few more MPH to become fully furled?

Hope this helps!

Have fun,

Scott.

[Flux]

Excellent Scott.

I only question one thing and that is the last paragraph. I think this is the region where seeking force comes into play in real life although our simple theory can't deal with it.

Up to an angle approaching 45deg to the wind the prop seems to want to stay upwind and this seems to produce a torque opposing that due to the thrust. At this point it does need a fair increase in wind speed to make it furl, but at some critical angle the seeking force seems to vanish and then it will furl easily. Most machines drop power significantly when fully furled, although the power extracted should be proportional to thrust and it should furl at constant power.

If the offset is too small you will never get it to furl even though the tail may be away from its stop by up to nearly 40 deg. I an sure this accounts for a lot of burn outs. People see this angle and assume it is furling, then the big wind comes and the furling occurs at a higher power out than they have seen before or expect.

Flux

[scottsAI]

Thanks Flux,

The last paragraph is weak, some things I have not figured out how to model let alone figured out what they are! Like why spinning blades will track with slow wind changes, with no tail.

Your comments make since to me, do not know how to incorporate into the model (physics) so I am stuck.

Have fun,

Scott.